Question

If the point of contact of the circle $x^2+y^2-30x+6y+109=0$ with the tangent 11x - 2y - 46 = 0 is (a, b), find a + b

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Answer 1

We can that the point of contact is the foot of the perpendicular of center c on the tangent. so, to find the point of contact, we just have to find the foot of the perpendicular of center on the tangent center c of the circle.

$x^2+y^2-30x+6y+109=0$ is (15,-3) we will find the foot of the perpendicular of (15,-3) on 11x-2y-46. It is given by

$\frac{x-15}{11}=\frac{y-(-3)}{-2}=\frac{-(11\times 15-2\times 3-46)}{{11}^2+2^2}$

=-1

$\Rightarrow x=4$ and y=-1