[[1]] is the equations of tangent and normal to the circle x2+y2−3x+4y−31=0 at the point (2,3).

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Detailed Solution

The given equation of the circle is
x2+y2−3x+4y−31=0 ....(1)
Let us denote the point as P(−2,3) and let the slope of tangent be mtmt.
We differentiate implicitly equation (1) with respect to x (x is the independent variable) to find mt,

\frac{d}{dx}

(x2+y2−3x+4y−31=0)=

\frac{d}{dx}

\Rightarrow 2 x + 3 y \frac{dy}{dx} - 3 + 4 \frac{dy}{dx} = 0

\Rightarrow (3 y + 4) \frac{dy}{dx} = 3 - 2 x

\Rightarrow \frac{dy}{dx} = \frac{3 - 2 x}{2 y + 4}

So the slope of the tangent at the point P(2,3) is

mt = \frac{3 - 4}{6 + 4} = - \frac{1}{10}

. We know from the point slope formula that the equation of line in a plane with point (x1,y1) and slope mm is given by

y - y 1 = m (x - x 1) . . . (2)

We put the values of mt , the point P(2,3) in equation(2) and get
y−(3)=mt(x−2)

\Rightarrow y - 3 = - \frac{1}{10} (x - 2)

⇒x+10y=32
We get find the equation of the tangent line to be x+10y=32
Let us denote the slope of normal as mn. The tangent line and the normal line are always orthogonal to each other. We know that the product of slopes of two orthogonal lines is −1. So, mtmn=−1

\Rightarrow (- \frac{1}{10}) mn = - 1

⇒mn=10
Now we shall use the point slope formula for the equation of line to find out the equation of normal. We put the values m=mn=10 and the point P(2,3) in equation(2) and get,
y−(3)=10(x−2)
⇒y−3=10(x−2)
⇒10x−y=17
So the obtained equation of normal is 10x−y=17.

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