Circles and System of Circles
(Geometry)
Engineering Entrance Exams
Study Material
Useful to 10th Class to 12th Class Students
Useful to Every Mathematics Student
Basically Targeted To All Engineering Entrance
Useful to Class-I Competitive Exams like
Civil Services Exams, Bank Probationary
Officer Exams and Staff Selection
Commission Exams
CONTENT OF THE STUDY MATERIAL
Circles Synopsis
Circles Problems with Solutions -1
Circles Problems with Solutions -2
System of Circles Synopsis
System of Circles Conceptual Theorems
Systems of Circles Problems with Solutions
By
MD.K.SHAREEF
20 Years Experience in Teaching
===================================================
Circles Synopsis
1. The locus of a point which moves such that its distance from a fixed
point is always constant is a circle. The fixed point is called centre
and the fixed distance is called radius of the circle.
2. Centre C = (a, b); point P = (x, y); radius CP = r
Equation of circle: (x - a)2 + (y - b)2 = r2
3. Centre C = (0, 0); point P = (x, y); radius CP = r
Equation of circle: x2 + y2 = r2
4. General equation of circle: x2 + y2 + 2gx + 2fy + c = 0
Notation: S = 0
Centre C = (-g, -f)
Radius =
5.
(x , y ), (x , y ) are the extremities of the diameter of the circle.
1 1
2 2
Equation: (x - x ) (x - x ) + (y - y ) (y - y ) = 0
1
2
1
2
6.
x - intercept: 2
7.
y - intercept: 2
8. Touches x - axis: s = ±
Touches y- axis: f = ±
9.
10. Touches x-axis at (x1, 0) and y- axis at (0, y1)
x2 + 2gx + e = (x - x1)2
y2 + 2fg + c = (y - y1)2
11. Notation
s ≡ x2 + y2 + 2gx + 2fy + c
s1 ≡ xx1 + yy1 + g(x + x1) + f(y + y1) + c
s2 ≡ xx2 + yy2 + g(x + x2) + f(y + y2) + c
s11 ≡ x12 + y12 + 2gx1 + 2fy1 + c
s12 ≡ x1x2 + y1y2 + g(x1+ x2) + f(y1+ y2) + c
12.
Length of the tangent:
Length of chord:
13.
14. Equation of Tangent at (x1, y1): xx1+ yy1 = r2 to the circle x2 + y2 = r2
15. Equation of tangent to the circle s = 0 is s1 = 0
(or) xx1+ yy1 + g(x + x1) + f(y + y1) + c = 0
16.Normal passes through the centre of the circle and normal is
perpendicular to the tangent.
17. Two tangents are drawn from a given external point to a given
circle.
18.
Chord of contact: Equation: s = 0
1
i.e., xx + yy + g(x + x ) + f(y + y ) + c = 0 to the circle s = 0
1
1
1
1
Equation: xx + yy = r to the circle x + y = r
1
1
19. Equation of chord joining A(x1, y1) and B(x2, y2) on the circle s = 0 is s1 + s2 = s12
20. Equation of chord with M(x1, y1) as its mid point and drawn to the circle s = 0 is s1 = s11
21. Angle between tangents:
=======================================================
Problems with Solutions of Circles-1
(1) Find the value of 'P', so that the points (2, 0), (0, 1), (4, 5) and (o, p) are
con cyclic?
Sol: Given Con cyclic points are; (2, 0), (0, 1), (4, 5) and (0, p)
Let the circle be: x2 + y2 + 2gx + 2fy + c = 0
At: (2, 0): 4 + 0 + 2g (2) + 2f (0) + c = 0
4g + c = -4
(1)
At (0, 1): 0 + 1 + 2g (0) + 2f (1) + c = 0
2f + c = -1
(2)
At (4, 5): 16 + 25 + 2g (4) + 2f (5) + c = 0
8g + 10f + c = - 41
(3)
(2) Find the equation of the circle which passes through (4, 1), (6, 5) and
whose centre lies on 4x + 3y - 24 = 0
Sol: Given points: (4, 1) and (6, 5)
Let the required circle be: x2 + y2 + 2gx + 2fy + c = 0
At (4, 1): 16 + 1 + 2g (4) + 2f (1) + c = 0
8g + 2f + c = - 17
(1)
At (6, 5) : 36 + 25 + 2g (6) + 2f (5) + c = 0
12g + 10f + c = - 61
(2)
Centre (-g, -f) lies on the line: 4x + 3y - 24 = 0
4 (-g) + 3 (-f) - 24 = 0
4g + 3f = - 24
(3)
(3) Show that the circles x2 + y2 - 4x - 6y - 12 = 0 and x2 + y2 + 6x + 18y + 26 = 0
touch each other. Find the point of contact and the equation of the common
tangent at the point of contact.
Sol: Given first circle; (S) = x2 + y2 - 4x - 6y - 12 = 0
Centre : C1 (2, 3).
Given second Circle; S' = x2 + y2 + 6x + 18y + 26 =0
Centre : C2 (-3, -9)
The given two circles touch each other externally
(4) Find the direct common tangents of x2 + y2 + 22x - 4y - 100 = 0 and
x2 + y2 - 22x + 4y + 100 = 0
Sol: Given first circle : (S) = x2 + y2 + 22x - 4y - 100 = 0
Centre : C1 (-11, 2)
Given second circle : (S') = x2 + y2 - 22x + 4y + 100 = 0
Centre : C2 (11, -2)
[xx1 + yy1 + 11 (x + x1) - 2 (y + y1) - 100]2 =(x2 + y2 + 22x - 4y - 100) (x12 + y12 +
22x1 - 4y1 - 100)
At (22,-4); [x (22) + y (-4) + 11 (x + 22) - 2 (y - 4) - 100]2 = (x2 + y2 + 22x - 4y
-100)
(484 + 16 + 22 (22) -4 (-4) - 100)
(33x - 6y + 150)2 = (x2 + y2 + 22x - 4y - 100) (900)
(11x - 2y + 50)2 = 100 (x2 + y2 + 22x - 4y - 100)
121x2 + 4y2 + 2500 - 44xy - 200y + 1100x - 100x2 - 100y2 - 2200x + 400y +
10000 = 0
21x2 - 44xy - 96y2 - 1100x + 200y + 12500 = 0
21x2 - 44xy - 96y2 = 21x2 + 28xy - 72xy - 96y2
= 7x (3x + 4y) -24y (3x + 4y)
21x2 - 44xy - 96y2 = (3x + 4y) (7x - 24y)
Now, 21x2 - 44xy - 96y2 - 1100x + 200y + 12500 = (3x + 4y + l) (7x - 24y + m)
Comparing x and y coefficients on both sides.
7l + 3m = -1100
(A)
-24l + 4m = 200
(B)
(5) Find the equation of the circle with centre (-2, 3) cutting a chord length of 2
units on 3x + 4y + 4 = 0.
Sol: Given Centre : C (-2, 3)
Given Line : 3x + 4y + 4 = 0
Required Circle : (x + 2)2 + (y - 3)2 = 5
x2 + y2 + 4x - 6y + 8 = 0
(6) Show that A (3, -1) lies on the circle x2 + y2 - 2x + 4y = 0 and find the other
end of diameter through A.
Sol: Given Circle ; x2 + y2 - 2x + 4y = 0
Centre : C (1, -2)
Given Point : A (3, -1)
substituting A (3, -1) in given circle
9 + 1 - 2 (3) + 4 (0) = 0
0=0
... A (3, -1) lies on the circle.
Other end of the diameter; B = (2C-A)
B - [(2, -4) - (3, -1)]
.
. . B = (-1, -3)
(7) Obtain the parametric equations of the circle x2 + y2 - 6x + 4y - 12 = 0
Sol: Given circle: x2 + y2 - 6x + 4y - 12 = 0
Centre : (3, -2)
=======================================================
Circles Problems with Solutions -2
1. The centre (C) and radius (r) of the circle 4x2 + 4y2 - 10x + 5y + 5 = 0 are
Ans: (2)
Explanation:
2. A circle has its centre at (a, b). If the radius is a + b, its equation is
1) x2 + y2 - ax - by - ab = 0
2) x2 + y2 + ax + by + ab = 0
3) x2 + y2 - 2ax - 2by - 2ab = 0
4) x2 + y2 + 2ax + 2by + 2ab = 0
Ans: (3)
Explanation: (x - a)2 + (y - b)2 = (a + b)2
x2 + y2 - 2ax - 2by + a2 + b2 = a2 + b2 + 2ab
... x2 + y2 - 2ax - 2by - 2ab = 0
3. Area of the triangle formed by the center of the circles
x2 + y2 = 1, x2 + y2 + 6x - 2y = 1, x2 + y2- 12x + 4y = 1 is
1) 0
2) 1
3) 3
4) 6
Ans: (1)
Explanation: A = (0, 0), B = (-3, 1), C = (6, -2)
4. A circle with centre at (2, 3) passes through (2, 4). Its equation is
1) x2 + y2 - 4x - 6y + 12 = 0
2) x2 + y2 - 6x - 4y + 12 = 0
3) x2 + y2 + 4x + 6y - 12 = 0
4) x2 + y2 + 6x + 4y - 12 = 0
Ans: (1)
Explanation: C = (2, 3)
Circle equation: (x - 2)2 + (y - 3)2 = 12
x2 + y2 - 4x - 6y + 4 + 9 = 1
x2 + y2 - 4x - 6y + 12 = 0
5. A circle with its centre at (4, 5) passes through the centre of the circle
x2 + y2 + 4x - 6y - 12 = 0. Its equation is
1) x2 + y2 + 8x + 10y + 1 = 0
2) x2 + y2 + 8x - 10y - 1 = 0
3) x2 + y2 - 8x + 10y - 1 = 0
4) x2 + y2 - 8x - 10y + 1 = 0
Ans: (4)
Explanation: Centre of the given circle = (-2, 3)
Centre of the required circle = (4, 5)
... Equation of the required circle:
(x - 4)2 + (y - 5)2 =
x2 + y2 - 8x - 10y + 41 - 40 = 0
x2 + y2 - 8x - 10y + 1 = 0
6. A circle centered at (2, 3) passes through the intersection of the lines 4x + y =
27 and
3x - 2y - 1 = 0. The radius of the circle is
1) 2
2) 3
3) 4
4) 5
Ans: (4)
Explanation: Point of intersection of 4x + y = 27 and 3x - 2y = 1 is (5, 7)
Centre of the circle = (2, 3)
7. A diameter of the circle x2 + y2 - 6x + 2y - 8 = 0 passes through origin. Its
equation is
1) x + 2y = 0
2) x - 2y = 0
3) x + 3y = 0
4) x - 3y = 0
Ans: (3)
Explanation: C = (3, -1)
O = (0, 0)
8. A circle is concentric with the circle x2 + y2 - 4x - 2y - 4 = 0. If it passes
through the center of the circle x2 + y2 + 2x + 4y = 0, its equation is
1) x2 + y2 - 4x - 2y + 13 = 0
2) x2 + y2 - 4x - 2y - 13 = 0
3) x2 + y2 - 4x - 2y + 12 = 0
4) x2 + y2 - 4x - 2y - 12 = 0
Ans: (2)
Explanation: Required circle: x2 + y2 - 4x - 2y + k = 0
Centre of given circle = (-1, -2)
... 1 + 4 + 4 + 4 = -k
... k = -13
... Equation required: x2 + y2 - 4x - 2y - 13 = 0
9. The equation of the circle concentric with the circle x2 + y2 - 6x + 12y + 15 = 0
and of double its radius is
1) x2 + y2 - 6x + 12y - 75 = 0
2) x2 + y2 - 6x + 12y + 75 = 0
3) x2 + y2 - 6x + 12y + 65 = 0
4) x2 + y2 - 6x + 12y - 65 = 0
Ans: (1)
Explanation:
10. A circle passes through (1, 1), (2, -1), (3, -2). Its radius is
Ans: (3)
Explanation:
... Equation of circle: x2 + y2 - 13x - 5y + 16 = 0
11. The equation of the circle passing through the points (1, 2), (3, 6), (5, -6) is
1) x2 + y2 + 22x - 4y + 15 = 0
2) x2 + y2 - 22x + 4y + 15 = 0
3) x2 + y2 + 22x + 4y + 25 = 0
4) x2 + y2 - 22x - 4y + 25 = 0
Ans: (4)
Explanation: 2g + 4f + c = -5; 6g + 12f + c = -45; 10g - 12f + c = -61
Solving g = -11, f = -2, c = 25
... Circle equation: x2 + y2 - 22x - 4y + 25 = 0
12. If a circle passes through (5, 7), (6, 6), (2, -2), its centre and radius are
1) (2, 3), 5
2) (3, 2), 5
3) (2, 5), 3
4) (3, 5), 2
Ans: (1)
Explanation: 10g + 14f + c = -74
12g + 12f + c = -72
4g - 4f + c = -8
g = -2, f = -3, c = -12
... C = (-g, -f) = (2, 3)
13. A circle has its centre on the line 4x + y - 16 = 0. If it passes through the points
(6, 5) and
(4, 1), its equation is
1) x2 + y2 - 6x + 8y + 25 = 0
2) x2 + y2 + 6x - 8y - 25 = 0
3) x2 + y2 - 6x - 8y + 15 = 0
4) x2 + y2 + 6x + 8y - 15 = 0
Ans: (3)
Explanation: 4(-g) + (-f) - 16 = 0 ...
(6, 5)
(4, 1)
Solving, g = -3, f = -4, c = 15
... Equation of circle: x2 + y2 - 6x - 8y + 15 = 0
14. A circle passing through (4, 2) and (-6, -2) has its centre on Y-axis. Its
equation is
1) x2 + y2 + 5y + 30 = 0
2) x2 + y2 + 5y - 30 = 0
3) x2 + y2 - 5y + 30 = 0
4) x2 + y2 - 5y - 30 = 0
Ans: (2)
Explanation: 8g + 4f + c = -20
-12g - 4f + c = -40
(-g, -f) lies on Y-axis g = 0
on solving, we get 2f = 5; c = -30
... Circle required: x2 + y2 + 5y - 30 = 0
15. A circle passes through the points (h, k), (7, 0), (5, 2) and (1, -6) then h + k + 5
=
1) -1
2) -4
3) -5
4) 0
Ans: (4)
Explanation:
g = -3, f = 2, c = -7
also (h, k) = (-1, -4)
h+k+5=0
16. The equation of the circle with radius 1 and passing through (1, 1) , (2, 2) is
1) x2 + y2 - 4x - 2y + 4 = 0
2) x2 + y2 + 4x + 2y + 4 = 0
3) x2 + y2 - 2x - 4y - 4 = 0
4) x2 + y2 + 2x + 4y - 4 = 0
Ans: (1)
Explanation:
g2 + f2 = 1 + c
2g + 2f + c = -2
4g + 4f + c = -8
solving, we get g = -2, f = -1, c = 4
... Circle equation: x2 + y2 - 4x - 2y + 4 = 0
17. A circle of 5 radius has its centre on X-axis. If it passes through (2, 3), its
equation is
1) x2 + y2 - 12x - 11 = 0
2) x2 + y2 + 12x + 11 = 0
3) x2 + y2 - 4x + 21 = 0
4) x2 + y2 + 4x - 21 = 0
Ans: (4)
Explanation:
5
g2 + f2 - c = 25
(-g, -f) on X-axis
f=0
(2, 3) 4g + 6f + c = -13
solving, we get g = 2, c = -21
... Equation of circle = x2 + y2 + 4x - 21 = 0
18. The equation to the locus of the point of intersection of the lines
x sin α - y cos α = b and x cos α + y sin α = a is
1) x2 + y2 = a2
2) x2 + y2 = b2 3) x2 + y2 = a2 + b2
4) xy = a2b2
Ans: (3)
Explanation: Squaring and adding, x2 + y2 = a2 + b2
19. The equation to the locus of the foot of the perpendicular from origin to the
line which always passes through a fixed point (h, k) is
1) x2 + y2 + hx + ky = 0
2) x2 + y2 + hx - ky = 0
3) x2 + y2 - hx + ky = 0
4) x2 + y2 - hx - ky = 0
Ans: (4)
Explanation:
Ans: (2)
Explanation:
21. A circle passing through origin makes positive intercepts a and b on the
coordinate axes. Its equation is
1) x2 + y2 + ax + by = 0
2) x2 + y2 + ax - by = 0
3) x2 + y2 - ax + by = 0
4) x2 + y2 - ax - by = 0
Ans: (4)
Explanation: When a circle makes positive intercepts a and b on the coordinate
axes, we have the end points of the diameter as (a, 0) and (0, b).
... The equation of the circle: (x - a)(x) + (y)(y - b) = 0
x2 + y2 - ax - by = 0
22. A circle circumscribes a square ABCD of side 'a'. If AB and AD represent the
coordinate axes, then the equation of the circle is
1) x2 + y2 + ax + ay = 0
2) x2 + y2 - ax - ay = 0
3) x2 + y2 + ax - ay = 0
4) x2 + y2 - ax + ay = 0
Ans: (2)
Explanation: When a circle circumscribes a square ABCD of side 'a' and AB and
AD represent the coordinate axes, we have A = (0, 0), B = (a, 0), C = (a, a), D =
(0, a).
... The equation of the circle with AC as diameter is
(x - 0)(x - a) + (y - 0)(y - a) = 0
x2 - ax + y2 - ay = 0
x2 + y2 - ax - ay = 0
23. A = (a, b), B = (c, d), P = (x, y) are three points. If
equation to the locus of P is
1) x2 + y2 - (a + c)x - (b + d)y + (ac + bd) = 0
2) x2 + y2 + (a + c)x + (b + d)y - (ac + bd) = 0
3) x2 + y2 - (a + b)x - (c + d)y + (ab + cd) = 0
4) x2 + y2 + (a + b)x + (c + d)y - (ab + cd) = 0
Ans: (1)
is a right angle, the
Explanation: (Slope of AP)(Slope of PB) = -1
(y - b)(y - d) + (x - a)(x - c) = 0
x2 + y2 - (a + c)x - (b + d)y + ac + bd = 0
24. (3, 4) is one end of diameter of a circle x2 + y2 - 4x - 6y + 11 = 0. The other
end is
1) (1, 2)
2) (2, 3)
3) (3, 1)
4) (3, 2)
Ans: (1)
Explanation: C = (2, 3) is the centre.
A = (3, 4) is one end of the diameter.
B = (h, k) is the other end
(h, k) = (1, 2)
1) ab(x + y) = a2 + b2
3) (bx + ay) = ab
Ans: (3)
Explanation:
2) (a2 + b2)(x + y) = a2b2
4) (ax + by) = ab
26. Equation of normal to 2x2 + 2y2 + 3x - 4y + 1 = 0 at (-1, 2) is
1) x + y + 2 = 0
2) 4x + y + 2 = 0
3) x + 4y + 2 = 0 4) 4x + 2y + 3 = 0
Ans: (2)
Explanation:
27. The equation of normal to the circle x2 + y2 - 8x - 2y + 12 = 0 at the points
whose ordinate -1 is
1) 2x - y -7 = 0
2) 2x - y + 7 = 0
3) 2x + y + 9 = 0
4) 2x - y - 9 = 0
Ans: (1)
Explanation: P = (h, -1) h2 + 1 - 8h + 2 + 12 = 0
h2 - 8h + 15 = 0
h = 3, 5
... P(3, -1) or (5, -1)
28. A line y = mx + c is a normal to the circle with centre (a, b) and radius r. Then
1) am = b + c
2) am = b - c
3) bm = a + c
4) bm = a - c
Ans: (2)
Explanation: Normal always passes through centre of a circle.
... y = mx + c passes through (a, b)
b = m(a) + c (or) am = b - c.
29. The normal at (2, 3) to the circle x2 + y2 + 4x + 6y - 39 = 0 meets the circle
also at
1) (-2, -3)
2) (-6, 9)
3) (6, -9)
4) (-6, -9)
Ans: (4)
Explanation: Normal passes through centre (-2, -3) of the circle.
Therefore, the normal becomes the diameter for which (2, 3) is one end.
If (h, k) is the other end, we have
... h = -6, k = -9
... Normal at (2, 3) meets the circle at (-6, -9).
30. A line parallel to x + 2y - 3 = 0 is a normal to the circle x2 + y2 - 2x = 0. Its
equation is
1) x + 2y - 1 = 0
2) x + 2y + 1 = 0
3) x + 2y + 2 = 0
4) x + 2y - 2 = 0
Ans: (1)
Explanation: x + 2y - k = 0 is parallel to x + 2y - 3 = 0
Passes through centre (1, 0) of the circle.
1 -k = 0
... k = 1
... The required equation is x + 2y - 1 = 0.
31. y = mx + c touches x2 + y2 = r2 at
Ans: (2)
Explanation: xx1 + yy1 = r2 tangent at (x1, y1)
mx - y = -c given tangent
32. y = x +
touches x2 + y2 = a2 at
Ans: (3)
Explanation:
33. The area of the triangle formed by the tangent at (α, β) to the circle x2 + y2 =
r2 with the coordinate axes is
Ans: (2)
Explanation:
34. 2b is the intercept made by the circle x2 + y2 = a2 with the line y = mx + c.
Then c2 =
1) a2(1 + m2)
2) b2(1 + m2)
3) (a2 + b2)(1 + m2)
4) (a2 - b2)(1 +
m2)
Ans: (4)
Explanation:
35. A line y = x + 2 is cut off by the circle x2 + y2 + 4x - 2y - 3 = 0. The middle
point of the line is:
Ans: (2)
Explanation: Middle point of the line is the foot of the perpendicular from the
centre of the circle.
=======================================================
System of Circles Synopsis
Definition:
System of Circles means a group of circles.
Radical Axis:
A line between two circles.
Radical centre:
The point of intersection of radical axes.
Coaxial system of circles: A system of circles is said to be coaxial when any
two circles of the system have the same radical axis. Any cirlcle belonging to the
coaxial system takes the form
where S = 0 is a member of the
system and L = 0 is the radical axis of the system. (
is a parameter)
Limiting points:
The members of a coaxial system of circles with zero radius are called the
limiting points of the coaxial system.
=======================================================
System of Circles Conceptual Theorems
1. If S = 0 and S' = 0 are two intersecting circles of radii r1 and r2 respectively, 'd'
is the distance between the two circles and 'θ' is the angle between the two circles
then
show that
Proof :
Let centers be : C1 and C2
Given : C1C2 = d
Given radii : r1 and r2
From figure: C1P = r1; C2P = r2
2. Show that the condition that the circles S = 0 and S' = 0 may cut each other
orthogonally is 2gg' + 2ff ' = c + c'.
Proof:
Given Circle (S)
x2 + y2 + 2gx + 2fy + c = 0
Centre : C1 ( -g, -f)
Radius :
Given circle (S') : x2 + y2 + 2g'x + 2f'y + c' = 0
Centre : C2 (-g', -f ')
Radius :
By data : C1 C2 = 90°
C1C22 = C1P2 + C2P2
(-g + g')2 + (-f + f ')2 = g2 + f2 - c + (g')2 + (f ')2 - c'
g2 + g'2 - 2gg' + f2 + f '2 - 2ff ' = g2 + f2 - c + g'2 + f '2 - c'
-2gg' - 2ff ' = -c - c'
3. Show that the equation to a system of coaxial circles in its simplest form is
x2 + y2 + 2λx + c = 0.
Let x2 + y2 + 2gx + 2fy + c = 0 represent the members of a coaxial system for
different values of g, f and c.
For the coaxial system, let the line of centres be the x-axis and the radical axis be
the y-axis.
Since the centres of all the circles of a coaxial system lie on the x-axis, their ycoordinates must be zero.
i.e. -f = 0
f=0
Let the equations of any two circles of the system be
x2 + y2 + 2g1x + c1 = 0 .......... (1)
x2 + y2 + 2g2x + c2 = 0 .......... (2)
The radical axis of (1) and (2) is
2 (g1 - g2) x + (c1 - c2) = 0
Since the radical axis is the y- axis i.e x = 0
c1 - c2 = 0
c1 = c2
Let c1 = c2 = c
then, the circles (1) and (2) become
x2 + y2 + 2g1x + c = 0 .......... (3)
x2 + y2 + 2g2 x + c = 0 .......... (4)
Similarly any other circle coaxial with
(3) and (4) will be x2 + y2 + 2g3x + c = 0
Hence, the equation of coaxial system of circles in simplest form is
x2 + y2 + 2λx + c = 0 where λ is a parameter and 'c' is a constant.
=======================================================
Systems of Circles Problems with
Solutions
1. Find the radical centre of the circles
x2 + y2 - 4x - 6y + 5 = 0, x2 + y2 - 2x- 4y- 1 = 0; x2 + y2 - 6x- 2y = 0.
Sol:
Given Circles: (S) : x2 + y2 - 4x - 6y + 5 = 0
(S') : x2 + y2 -2x-4y-1 = 0
(S'') : x2 + y2- 6x- 2y = 0
S - S' = 0
-2x -2y + 6 = 0
x + y -3 = 0 ................. (1)
S'- S'' = 0
4x -2y-1 = 0 ................. (2)
Solving (1) and (2) we get the radical centre.
2. If the circles x2 + y2 + 2ax + c = 0 and x2 + y2 + 2bx + c = 0 touch each other
then show that
.
Sol: Given first Circle : x2 + y2 + 2ax + c = 0
Centre : C1(-a, 0)
By data : C1C2 = r1 ± r2
3. Find the equation of the circle Coaxial with the circles
x2 + y2 - 2x + 2y + 1 = 0 and x2 + y2 + 8x - 6y = 0 which passes through
the point (-1, -2) .
Sol: Given circles: (S) : x2 + y2 -2x + 2y + 1 = 0 (S' ) : x2 +y2 + 8x- 6y = 0
Radical axis (L) :
-10x + 8y + 1 = 0
10x - 8y - 1 = 0
Any circle of coaxial system of circle : S + λL =0
x2 + y2-2x + 2y + 1 + λ (10x- 8y-1) = 0
x2 + y2 + (10λ- 2) x + (2– 8λ) y+ (1-λ) = 0
Since this circle passes through (-1, -2)
1 + 4 -1 (10λ-2) -2 (2-8λ) + (1-λ) = 0
5-10λ + 2 -4 + 16λ + 1- λ = 0
5λ + 4 = 0
Required circle : x2 + y2 - 2x + 2y + 1 - ( 10x - 8y -1) = 0
5(x2 + y2) - 10x +10y +5 - 40x + 32y + 4 = 0
5(x2 + y2) - 50x + 42y + 9 = 0
4. Find the equation of the circle which is coaxial with the circles
x2 + y2 -6x + 4 = 0 and x2 + y2 -5x + 4 = 0 and touches the line 3x - 4y = 15 .
Sol: Given circles:
(S) : x2 + y2 -6x + 4 = 0
(S') : x2 + y2 -5x + 4 = 0
Radical axis (L) :
-x = 0
x=0
Any circle of coaxial system of
circle : S + λL = 0
x2 + y2 -6x + 4 + λ (x) = 0
x2 + y2 + x (λ-6) + 4 = 0
Given line: 3x- 4y-15 = 0
Squaring on both sides
9λ2 + 144 + 72λ = 25λ2 - 300λ + 500
16λ2 -372λ + 356 = 0
4λ2 - 93λ + 89 = 0
4λ2 - 4λ - 89λ + 89 = 0
4λ(λ -1) - 89 (λ -1) = 0
(λ -1) (4λ - 89) = 0
... λ = 1;
Required circles:
λ=1
x2
2
2 -5x + 4 = 0
x +y
x2
2
+ y -6x + 4 + 1 (x) = 0
2 -6x + 4 +
+y
(x) = 0
2
2
4( x + y ) - 24x + 16 + 89x = 0
2
2) + 65x + 16 = 0
4( x + y
5. Find the angle between the circles x2 + y2 + 4x -14y + 28 = 0
and x2 + y2 + 4x - 5 = 0.
Sol: Given first circle : (S) : x2 + y2 + 4x -14y + 28 = 0 ; Centre : C1 (-2, 7)
... θ = 60°
6. Find the equation to the circle whose diameter is the common chord of the two
circles x2 + y2 - 4x + 6y -12 = 0 and x2 + y2 + 2x - 2y - 23 = 0. Hence find the
length of the common chord.
Sol: Given Circles : (S) : x2 + y2 - 4x + 6y-12 = 0 ; (S' ) : x2 + y2 + 2x- 2y- 23 = 0
Common chord (L) :
-6x + 8y + 11 = 0
6x - 8y - 11 = 0
Any circle of coaxial system of circle: S + λL=0
x2 + y2 - 4x + 6y-12 + λ (6x - 8y- 11) = 0
x2 + y2 + x(6λ - 4) + y (6 - 8λ) + (-12 - 11λ)
=0
Centre : (2 - 3λ , 4λ - 3)
By data : 6(2 - 3λ) -8 (4λ-3) -11 = 0
12 -18λ - 32λ + 24 - 11 = 0
-50λ + 25 = 0
Required Circle: x2 + y2 - 4x + 6y - 12 + (6x - 8y - 11) = 0
2(x2 + y2) - 8x + 12y - 24 + 6x - 8y - 11 = 0
x2 + y2 - x + 2y 2g = -1 ;
2f = 2 ;
g=- ;
= 0
C=f = 1
Length of the Common Chord : 2(Radius)
7. Find the equation of the circle which is orthogonal to the circles
x2 + y2 + 2x +17y + 4 = 0, x2 + y2 + 7x + 6y + 11 = 0 and x2 + y2 - x + 22y + 3 = 0.
Sol: Let the required circle be : x2 + y2 + 2gx + 2fy + c = 0 .
Given Circle : x2 + y2 + 2x + 17y + 4 = 0
g' = 1;
f'' =
;
c' = 4
Orthogonal Condition :
2g(1) + 2f ( ) = c + 4
2g + 17 f - c = 4 ................ (1)
Given Circle : x2 + y2 + 7x + 6y + 11 = 0
g' = ;
f'=3;
c' = 11
Orthogonal condition : 2g ( ) + 2f(3) = c + 11
7g + 6f - c = 11 ................ (2)
2
Given Circle : x + y2 - x + 22y + 3 = 0
g' = -
;
f'' = 11;
c' = 3
Orthogonal condition : 2g (- ) + 2f (11) = c + 3
-g + 22f - c = 3 ................ (3)
from (1) = 2(-3) + 17(-2) - c = 4
- 40 - c = 4
c = - 44
Required circle : x2 + y2 + 2(-3) x + 2(-2) y - 44 = 0
x2 + y2 - 6x - 4y - 44 = 0
8. Find the equation of the circle which passes through the point (0, -3) and
intersects the circles given by the equations x2 + y2 - 6x + 3y + 5 = 0 and x2 + y2 x- 7y = 0 orthogonally.
Sol: Let the required circle be: x2 + y2 + 2gx + 2fy + c = 0
Since this circle passes through (0, -3)
0 + 9 + 2g(0) + 2f(-3) + c = 0
-6f + c = -9 .................. (1)
2
Given Circle : x + y2 - 6x + 3y + 5 = 0
g' = -3 ;
f'' = ;
c' = 5
Orthogonal Condition :
2g(-3) + 2f ( ) = c + 5
-6g + 3 f - c = 5 .................. (2)
Given circle : x2 + y2 - x - 7 y = 0
g' = - ;
f'=- ;
c' = 0
Orthogonal condition : 2g (- ) + 2f (- ) = c + 0
-g -7 f- c = 0
g + 7 f + c = 0 .................. (3)
Solving (1) and (2)
Solving (4) and (5)
6g + 3f = 4
6g - 12f = -6
- +
+
3 (x2 + y2) + 2x + 4y -15 = 0
9. Find the equation of the circle which pass through the origin, having its centre
on the line x + y = 4 and intersecting the circle x2 + y2 - 4x + 2y + 4 = 0
orthogonally.
Sol : Let the required Circle be : x2 + y2 + 2gx + 2fy + c = 0
Since this Circle passes through (0, 0)
Centre (-g, -f) lies on the line : x + y = 4
-g - f = 4
g + f = - 4 .................. (1)
Given Circle : x2 + y2 - 4x + 2y + 4 = 0
g' = -2 ;
f'' = 1 ;
c' = 4
Orthogonal Condition :
2g(-2) + 2f(1) = 0 + 4
- 4g + 2f = 4
2g - f = -2 .................. (2)
Solving (1) and (2)
g+ f= -4
2g - f = - 2
3g
= -6
from (1) : -2 + f = -4
Required circle : x2 + y2 + 2(-2) x + 2(-2) y + 0 = 0
x2 + y2 - 4x - 4y = 0
10. Find the limiting points of the coaxial system, determined by the circles x2 +
y2 + 10x - 4y- 1 = 0 and x2 + y2 + 5x + y + 4 = 0.
Sol: Given circle (S) : x2 + y2 + 10x - 4y- 1 = 0
(S') : x2 + y2 + 5x + y + 4 = 0
Radical axis (L) : S - S' = 0
5x - 5y - 5 = 0
x-y-1=0
Any circle of coaxial system of circle : S + λ L = 0
x2 + y2 + 10x - 4y- 1+ λ (x- y- 1) = 0
x2 + y2 + (λ + 10) x + (- 4-λ) y + (-1 - λ) = 0
λ2+ 100 + 20λ + λ2 + 16 + 8λ + 4 + 4λ = 0
2λ2 + 32λ + 120 = 0
λ2 + 16λ + 60 = 0
λ2 + 6λ + 10λ + 60 = 0
λ(λ + 6) + 10 (λ + 6) = 0
(λ + 6) (λ + 10) = 0
λ = -6; λ = -10
Substituting the values of λ in centre, we get
Limiting points : λ = -6
(-2, -1)
λ = -10
(0, -3)
11. The point (2,1) is a limiting point of a coaxial system of circle of which
x2 + y2 - 6x - 4y - 3 = 0 is a member. Find the radical axis and the other limiting
point.
Sol: Given limiting point : (2,1)
Then circle (S) : (x- 2)2 + (y- 1)2 = 0
x2 + y2 - 4x- 2y + 5 = 0
Given Circle (S') : x2 + y2 - 6x - 4y - 3 = 0
Radical axis (L) ; S - S' = 0
2x + 2y + 8 = 0
x+y+4=0
Let Other limiting point be: (h, k) ; Then
h - 2 = k - 1 = -7
h - 2 = -7 ;
k -1 = -7
h = -5 ;
k = -6
... Other limiting point = (-5, -6)
12. The origin is a limiting point of a coaxial system of which
x2 + y2 + 2gx + 2fy + c =0 is a member prove that other limiting point is
.
Sol: Given Limiting Point: (0, 0) ; Then circle (S) : (x- 0)2 + (y- 0)2 = 0
x2 + y2 = 0
Given circle (S') : x2 + y2 + 2gx + 2fy + c = 0
Radical axis (L) : S - S' = 0
- 2gx- 2fy- c = 0
2gx + 2fy + c = 0
Any circle of coaxial system of circle : S + λL = 0
x2 + y2 + λ(2gx + 2fy+ c) = 0
x2 + y2 + 2gλx + 2fλy + cλ = 0
Centre : (- gλ, - fλ)
Radius :
By data : Radius = 0
g2λ2 + f2λ2 - cλ = 0
λ2 (g2 + f2) - cλ = 0
λ [ λ (g2 + f2) - c] = 0
λ = 0; λ (g2 + f2) - c = 0
13. Find the equation of the circle which passes through the origin and which
belongs to the coaxial system of which the liming points are (1, 2) and (4, 3).
Sol: Given Limiting Points : (1, 2) and (4, 3)
From (1, 2) : Circle (S) : (x- 1)2 + (y- 2)2 = 0
2
x + y2 - 2x - 4y + 5 = 0 .................. (1)
From (4, 3) : (S' ) (x - 4)2 + (y - 3)2 = 0
x2 + y2 - 8x - 6y + 25 = 0 .................. ( 2)
Radical Axis (L) : S - S' = 0
6x + 2y - 20 = 0
3x + y - 10 = 0
Any circle of coaxial system of circle : S + λL = 0
x2 + y2 - 2x- 4y + 5 + λ ( 3x + y- 10) = 0
Since it passes through ( 0, 0 )
5 - 10λ = 0
Required Circle : x2 + y2 - 2x - 4y + 5 + (3x +y - 10) = 0
2 (x2 + y2) - 4x - 8y + 10 + 3x + y - 10 = 0
2 (x2 + y2) - x - 7y = 0
14. (Find the equation of the circle which belongs to the coaxial system
determined by the limiting points (0, - 3) and (- 2, - 1) and which is orthogonal to
the circle x2 + y2 + 2x + 6y + 1 = 0.
Sol:
Given Limiting Points : (0, -3) and (-2, -1)
From (0, - 3) : Circle : (S) : (x- 0)2 + (y + 3)2 = 0
x2 + y2 + 6y + 9 = 0 .................. (1)
From (- 2, - 1) , Circle : (S' ) : (x + 2)2 + (y +1)2 = 0
x2 + y2 + 4x + 2y + 5 = 0 .................. ( 2)
Radical Axis (L) : S- S' = 0
- 4x + 4y + 4 = 0
x-y-1=0
Coaxial System of Circle : S + λL = 0
x2 + y2 + 6y + 9 + λ (x - y- 1) = 0
x2 + y2 + λx + (6 - λ) y + (9 - λ) = 0 .................. (3)
Given Circle : x2 + y2 + 2x + 6y + 1 = 0 .................. (4)
g' = 1;
f'' = 3;
c' = 1
Orthogonal Condition for (3) and (4) :
λ + 18 - 3λ = 10 - λ
18 - 2λ = 10 - λ
Required Circle : x2 + y2 + 6y + 9 + 8 (x - y - 1) = 0
x2 + y2 + 6y + 9 + 8x - 8y - 8 = 0
x2 + y2 + 8x - 2y + 1 = 0
15. Find the equation to the system of circle orthogonal to the coaxial system
x2 + y2 + 3x + 4y - 2 + λ (x + y - 7) = 0.
Sol:
Equation to given coaxial System is
x2 + y2 + 3x + 4y - 2 + λ (x + y - 7) = 0
x2 + y2 + x (3 + λ) + y(4 + λ) + (- 2 - 7λ) = 0 .................. (1)
Let the Equation to the Circle Orthogonal to the given Coaxial
System be : x2 + y2 + 2gx + 2fy + c = 0 .................. ( 2 )
By data: (1) and (2) are Orthogonal
g (3 + λ) + f (4 + λ) = c - 2 - 7λ
λ (g + f + 7) + 3g + 4f - c + 2 = 0
λ
g + f + 7 = 0 .................. ( 3 )
3g + 4f + 2 - c = 0 .................. ( 4 )
solving (3) and (4)
g = - 26 - c; f = 19 + c
Equation (2) becomes
x2 + y2 + 2 (- 26 - c) x + 2(19 + c) y + c = 0
x2 + y2 - 52x + 38y - c (2x - 2y - 1) = 0
Which is the required orthogonal system, when 'C' is a parameter.
16. Prove that the limiting points are inverse points with respect to every circle
of
coaxial system.
Sol: Let the equation of a circle of a coaxial system be : x2 + y2 + 2λx + c = 0.
Centre : C ( - λ , 0)
Radius :
By data : Radius = 0
λ2 - c = 0
λ2 = c
L (-
, 0) and L' (
, 0) are Limiting points
CL = - λ +
CL' = - λ CL . CL' = (- λ + ) (- λ - )
CL . CL' = λ2 - c
CL . CL' = r2
also C, L, L' are Collinear and L, L' lie on the same circle of C
Hence, L and L' are Inverse points w.r.t. every Circle of the system.
17. Prove that the equation of the circle which cuts orthogonally each member of
a given coaxial system x2 + y2 + 2 λx + c = 0 is x2 + y2 + 2fy - c = 0.
Sol:
The given coaxial system is given by the equation :
2
2
x + y + 2λx + c = 0 .................. (1)
Consider tow circle of this System for values λ1 and λ2 of λ.
x2 + y2 + 2λ1x + c = 0 .................. (2)
x2 + y2 + 2λ2x + c = 0 .................. (3)
x2 + y2 + 2gx + 2fy + k = 0
Let the Circle Orthogonal to (2) and (3) be ... 2 λ1g = c + k ; 2λ2g = c+k
2 (λ1 - λ2) g = 0
g = 0 [... λ1 ≠ λ2]
.
.. k = - c
Hence the equation of the circle which cuts every member of the given
coaxial system orthogonally is x2 + y2 + 2fy - c = 0 .
18. Show that the equation x2 + y2 + 2 (1 - λ) x + 2 (1 - 2λ) y - 2 = 0 where λ is a
parameter, represents a system of Coaxial Circles. Also find the equation of the
circle which cuts orthogonally the circlex2 + y2 + 4x - 6y + 8 = 0.
Sol: Given equation: x2 + y2 + 2 (1 - λ) x + 2 (1 - 2λ) y - 2 = 0
x2 + y2 + 2x + 2y - 2 - λ(2x + 4y) = 0
This clearly represents a circle
Let λ1 and λ2 be two values of the parameter λ, then the two circles
corresponding to two values of λ are x2 + y2 + 2x + 2y - 2 - λ1 ( 2x + 4y ) =
0 .................. (2)
and
x2 + y2 + 2x + 2y - 2 - λ2 ( 2x + 4y) = 0 .................. (3)
Radical axis of (2) and (3) : ( λ1 - λ2 ) ( 2x + 4y ) = 0
2x + 4y = 0 [... λ1 ≠ λ2]
That is every pair of circle S + λL = 0 has the same radical axis 2x + 4y = 0
Equation (1) represents a coaxial system of circles
Given another circle : x2 + y2 + 4x - 6y + 8 = 0 .................. (4)
(1) and (4) cut each other orthogonally.
2 (1 - λ) (2) + 2 (1 - 2 λ) ( - 3) = - 2 + 8
2- 2 λ - 6 + 12 λ = 6
10 λ - 4 = 6
10 λ = 10
λ=1
The equation of the circle cutting (4) orthogonally and belonging to (1) is
x2 + y2 + 2x + 2y - 2 - ( 2x + 4y ) = 0
2
2
x + y - 2y- 2 = 0
19. (- 2, - 1) is a limiting point of a coaxial system of which
x2 + y2 + 2x + 4y + 7 = 0 is a member find the equation of the orthogonal system.
Sol:
Given Limiting point: (- 2, - 1)
From (- 2, - 1) : Circle (S) : (x + 2)2 + (y + 1)2 = 0
x2 + y2 + 4x + 2y + 5 = 0 .................. (1)
Given circle (S' ) : x2 + y2 + 2x + 4y + 7 = 0 .................. (2)
Radical axis of (1) and (2) : (L) : 2x - 2y - 2 = 0
x-y-1=0
Coaxial System of Circle : S + λL = 0
x2 + y2 + 4x + 2y + 5 + λ ( x - y - 1 ) = 0
x2 + y2 + ( 4 + λ ) x + ( 2 - λ ) y + ( 5 - λ ) = 0
16 + λ2 + 8 λ + λ2 + 4 - 4 λ - 20 + 4 λ = 0
2λ2 + 8λ = 0
λ2 + 4λ = 0
λ [λ + 4] = 0
λ = 0; λ = - 4
Limiting points : λ = 0
(- 2, - 1)
λ= -4
(0, - 3)
Now any circle through the limiting points be x2 + y2 + 2gx + 2fy + c = 0
Where 4 +1 - 4g - 2f + c = 0 and 0 + 9 + 0 - 6f + c = 0
- 4g - 2f + c + 5 = 0 and - 6f + c + 9 = 0
Solving - 4g - 2f + c + 5 = 0
- 6f + c + 9 = 0