10-2
Circles
A tangent is a line that intersects the circle at
exactly one point. A tangent to a circle is
perpendicular to the radius at the point of tangency.
Tangent line
Holt Algebra 2
10-2
Circles
Write the equation of the line tangent to the
circle x2 + y2 = 29 at the point (2, 5).
Step 1 Identify the
center and radius of the
circle.
From the equation x2 + y2 = 29,
the circle has center of
(0, 0) and radius r =
.
Holt Algebra 2
10-2
Circles
Example 4 Continued
Step 2 Find the slope of the radius at the point of
tangency and the slope of the tangent.
Use the slope formula.
Substitute (2, 5) for (x2 , y2 )
and (0, 0) for (x1 , y1 ).
The slope of the radius is
5
2
.
Because the slopes of perpendicular lines are
negative reciprocals, the slope of the tangent is –
Holt Algebra 2
2
5
.
10-2
Circles
Example 4 Continued
Step 3 Find the slope-intercept equation of the
tangent by using the point (2, 5) and the slope
m = – 25 .
Use the point-slope formula.
2
Substitute (2, 5) (x1 , y1 ) and – 5 for m.
Rewrite in slope-intercept form.
Holt Algebra 2
10-2
Circles
Example 4 Continued
The equation of the line that is tangent to
x2 + y2 = 29 at (2, 5) is
.
Check Graph the
circle and the line.
Holt Algebra 2
10-2
Circles
Write the equation of the line that is tangent
2
2
to the circle 25 = (x – 1) + (y + 2) , at the
point (5, –5).
Step 1 Identify the center and radius of the circle.
From the equation 25 = (x – 1)2 +(y + 2)2, the
circle has center of (1, –2) and radius r = 5.
Holt Algebra 2
10-2
Circles
Check It Out! Example 4 Continued
Step 2 Find the slope of the radius at the point of
tangency and the slope of the tangent.
Use the slope formula.
Substitute (5, –5) for (x2 , y2 )
and (1, –2) for (x1 , y1 ).
The slope of the radius is
–3
4
.
Because the slopes of perpendicular lines are
negative reciprocals, the slope of the tangent is
Holt Algebra 2
.
10-2
Circles
Check It Out! Example 4 Continued
Step 3. Find the slope-intercept equation of the
tangent by using the point (5, –5) and the slope
.
Use the point-slope formula.
Substitute (5, –5 ) for (x1 , y1 ) and
for m.
Rewrite in slope-intercept form.
Holt Algebra 2
4
3
10-2
Circles
Check It Out! Example 4 Continued
The equation of the line that is tangent to 25 =
(x – 1)2 + (y + 2)2 at (5, –5) is
.
Check Graph the
circle and the line.
Holt Algebra 2
10-2
Circles
HW pg. 732
#’s 10, 11, 19 – 21, 31, 32
Holt Algebra 2