Question

Find the equation of tangent and normal to the circle x²+y²-3x+4y-31=0 at the point (-2,3).​

Answer 1

Step-by-step explanation:

${x}^{2} + {y}^{2} - 3x + 4y - 31 = 0 \\ \\ putting \: ( - 2) \: in \: x \: and \: 3 \: in \: y \\ \\ {( - 2)}^{2} + {3}^{2} - 3 \times ( - 2) + 4 \times 3 - 31 = 0 \\ \\ 4 + 9 + 6 + 12 - 31 = 0 \\ \\ 31 - 31 = 0 \\ \\ 0 = 0$

Answer 2

The equation of tangent at point (-2,3)  to the given circle

$7x-10y+44=0$

The equation of normal at point (-2,3) to the given circle is given by

$10x+7y=1$

Step-by-step explanation:

Given equation:

$x^2+y^2-3x+4y-31=0$

Differentiate w.r.t x

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

$2yy'+4y'=3-2x$

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

$y'=\frac{3-2x}{2y+4}$

Substitute x=-2 and y=3

$y'=\frac{3-2(-2)}{2(3)+4}=\frac{7}{10}$

$m=y'=\frac{7}{10}$

Point-slope form:

$y-y_1=m(x-x_1)$

Using the formula

The equation of tangent at point (-2,3)  to the given circle

$y-3=\frac{7}{10}(x+2)$

$10y-30=7x+14$

$7x-10y+14+30=0$

$7x-10y+44=0$

Slope of normal=$-\frac{1}{m}=-\frac{10}{7}$

The equation of normal at point (-2,3) is given by

$y-3=-\frac{10}{7}(x+2)$

$7y-21=-10x-20$

$10x+7y=-20+21=1$

$10x+7y=1$

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