Question

find the equation of tangent and normal at (1, 1) of the circle x^2+y^2-2x-5y+3=0

Answer 1

Hiii......friend,

Here is ur answer,

Tangent to the circle is S1 =0.

$= > x(1) + y(1) - (x + 1) - \frac{5}{2} (y + 1) + 3 = 0$

$= > 2y - 2 - 5y - 5 + 3 = 0$

$= > y = \frac{ - 4}{3}$

Normal at (1,1)

Since tangent & normal are perpendicular at a point (1,1)

=> So, Normal equation will be X= k.

Here it passes through (1,1) . So,

=> X= 1. is the normal to the circle.

:-) Hope it helps u.