Question

find the equationof tangent and normal to the circle x²+y²-3x+4y-31=0 at the point (-2;3)​

Answer 1

Answer:

equation of tangent at any point (h,k) on a circle is given by

$xh + yk + (x + h) + (y + k) + c = 0$

here (h,k) is (-2,3)

then equation of the tangent is

$- 2x + 3y + (x - 2) + (y + 3) - 31 = 0 \\ - x + 2y - 32 = 0 \\ x - 2y + 32 = 0.$

now, for the normal find the slope of the tangent.

here the slope is (1/2).

hence slope of the normal is -2.

therefore equation of the normal can be obtained by

$(y - 3) = ( - 2)(x + 2) \\ (y - 3) = - 2x - 4 \\ 2x + y + 1 = 0.$