Question

Find equation of tangent to the circle x² + y²-4x+3y +2=0 at the point (4,-2)​

Answer 1

Step-by-step explanation:

we can use calculus for this problem

to find the slope of the tangent, let us take derivative.of the equation of the circle

on differentiation, we get

2x + 2y y' - 4 + 3 y' + 0 = 0

=> 2x + 2y y' - 4 + 3 y' = 0

=> 2y y' + 3 y' = 4 - 2x

=> y' (2y + 3) = 4 - 2x

=> y' = (4 - 2x)/(2y + 3)

therefore at (4, -2), y' = (4 - 2*4)/(2*(-2) + 3)

=> y' = -4/-1

=> y' = 4

y' = m = 4

using slope point form, equation of the tangent

=> (y - y1) = m (x - x1)

=> y - (-2) = 4(x - 4)

=> y + 2 = 4x - 16

=> 4x - y - 18 = 0 is the equation of the tangent to tne

given circle at (4, -2)