Condition for the line y
= mx + c to
be a tangent to the circle x2
+ y2 = a2 and
finding the point of contact
Let the line y = mx + c touch the circle x2
+ y2 = a2 . The centre and radius of the
circle x2 + y2
= a2 are (0, 0) and a respectively.
Then the perpendicular distance of the line y − mx −
c = 0 from (0, 0) is
This must be equal to radius .Therefore = a or c2 = a2 (1+ m2) .
Thus the condition for the line y = mx + c
to
be a tangent to the circle x2 + y2
= a2 is c2 = a2
(1 + m2).
Let (x1 , y1 ) be the the
point of contact of y = mx + c with the circle x2
+ y2 = a2,
Then y1 = mx1 + c ……….(1)
Equation of tangent at (x1 , y1
) is xx1 +
yy1 = a2
.
yy1 = -xx1 + a2 ... (2)
Equations (1) and (2) represent the same line and hence the coefficients are proportional.
Then the points of contact is either
Note
The equation of tangent at P to a circle is y = mx ±
a √[1 +m2]
Theorem 5.4
From any point outside the circle x2 + y2
= a2 two tangents can be drawn.
Proof
Let P(x1 , y1 ) be a point outside the
circle. The equation of the tangent is
It passes through (x1, y1) . Therefore
Squaring both sides, we get
( y1 – mx1)2 = a2 (1+ m2 )
y12 + m2 x12 - 2mx1y1 - a2
- a2m2 = 0
m2 (x12
- a2 ) - 2mx1y1
+ ( y12
- a2 ) = 0 .
This quadratic equation in m gives two values for m .
These values give two tangents to the circle x2
+ y2 = a2.
Note
(1) If (x1, y1 ) is a point outside the circle, then both the
tangents are real.
(2) If (x1, y1 ) is a
point inside the circle, then both the tangents are imaginary.
(3) If (x1, y1 ) is a
point on the circle, then both the tangents coincide.
Example 5.11
Find the equations of the tangent and normal to the circle x2
+ y2 = 25 at P(−3, 4) .
Solution
Equation of tangent to the circle at P(x1
, y1 ) is xx1 + yy1 = a2
That is, x(-3) + y(4) = 25
-3x + 4 y = 25
Equation of normal is xy1 - yx1
= 0
That is, 4x + 3y = 0 .
Example 5.12
If y = 4x + c is a tangent to the circle x2
+ y2 = 9 , find c.
Solution
The condition for the line y = mx + c to be
a tangent to the circle x2 + y2 = a2
is c2 = a2 (1+ m2).
Then, c = ± √[9(1+16)]
c = ±3 √17.
Example 5.13
A road bridge over an irrigation canal have two semi circular
vents each with a span of 20m and the supporting pillars of width 2m .
Use Fig.5.16 to write the equations that represent the semi-verticular vents.
Solution
Let O1 O2 be
the centres of the two semi circular vents.
First vent with centre O1 (12, 0) and radius r = 10 yields equation to first semicircle as
(x -12)2 + ( y - 0)2 = 102
Þ x2
+ y2 - 24x + 44 = 0 , y > 0 .
Second vent with centre O2 (34, 0) and radius r
= 10 yields equation to second vent as
(x - 34)2 + y2 = 102
Þ x2
+ y2 - 68x +1056 = 0 , y > 0 .
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