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Example 5: Find a joint equation to the pair of tangents drawn from (5, 0) to the circle:
x2 + y2 = 9 (1)

Solution: Let P(h,k) be any point on either of the two tangents drawn from A(5,0) to the
given circle (1). Equation of PA is

y=- 0 k - 0 (x - 5) or kx - (h - 5) y =- 5k 0 (2)
S ince (2) is tahn-g5ent to the circle (1), the perpen dic ula r distance of (2) from the centre of the

circle equals the radius of the circle.

i.e., -5k = 3
k 2 + (h - 5)2

or 25k 2= 9[k 2 + (h - 5)2 ] or 16k 2 - 9(h - 5)2 = 0

Thus (h,k) lies on

9(x - 5)2 - 16y2 = 0 (3)

But (h,k) is any point of either of the two tangents.

Hence (3) is the joint equations of the two tangents.

6.2.1 Length of the tangent to a circle
(Tangential Distance)

Let P(x1, y1) be a point outside the circle
x2 + y2 +2gx + 2fy + c = 0 (1)
We know that two real and distinct tangents can be drawn to the circle from an external
point P. If the points of contact of these tangents with the circle are S and T, then each of
the length PS and PT is called length of the tangent or tangential distance from P to the
circle (1).
The centre of the circle has coordinates
(-g, -f). Join PO and OT. From the right triangle OPT
we have,

length of the tangent ==PT OP2 - OT 2

= (x1 + g)2 + ( y1 + f )2 - (g 2 + f 2 - c)

version: 1.1

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