Question

Find the equation of a line passing through $\left(-2,3\right)$ and parallel to tangent at origin for the circle $x^2+y^2+x-y=0$.

Answer 1

Given equation of circle :- $x^2+y^2+x-y=0$

Slope of tangent = $\frac{dy}{dx}$

Differentiating the equation w.r.t. x, $2x+2y\frac{dy}{dx}+1-\frac{dy}{dx}=0$

$\Rightarrow (1-2y)\frac{dy}{dx}=1+2x$

$\Rightarrow \frac{dy}{dx}=\frac{1+2x}{1-2y}$

So, Slope of tangent at origin = $\frac{dy}{dx}$ at origin = $\frac{1+2\times 0}{1-2\times 0}=\frac{1}{1}=1$

Since slope of parallel lines are equal.

Therefore, slope of line parallel to tangent at origin = 1

Point through which this line passes $\equiv (-2,3)$

Thus, point-slope form of equation is $y-y_1=m(x-x_1)$

$\Rightarrow y-3=1\left[x-(-2)\right]\Rightarrow y-3=x+2\Rightarrow y=x+5$