Question

Find the equation of the normal to the circle $x^2+y^2-6x-8y-24=0$ at the point (4, 6).

• A. 2x -y = 2
• B. 3x + y = 18
• C. 3x - y = 12
• D. -3y + 12 = 0

Answer 1

The correct option is A

2x -y = 2

$L_1$ is the tangent at P. the normal at P is the line which is perpendicular to tangent and passes through P.

we can see that it passes through center of circle.

so the equation of normal can be obtained by using center and point of contact

$\Rightarrow$ Normal is the straight line passing through P(4,6) and C(3,4)

$\Rightarrow y-4=\frac{6-4}{4-3}(x-3)$

$\Rightarrow y-4=2x-6$

$\Rightarrow 2x-y=2$