Question

The tangent to the circle $x^2+y^2=5$ at the point $\left(1,-2\right)$ also touches the circle $x^2+y^2-8x+6y+20=0$ at

• A. $\left(-2,1\right)$
• B. $\left(-3,0\right)$
• C. $\left(-1,-1\right)$
• D. $\left(3,-1\right)$

Answer 1

The correct option is D $\left(3,-1\right)$

for tangent

Differentiating equation $x^2+y^2=5,$ we get,

$2x+2y\frac{dy}{dx}=0$

therefore , at$(1,-2)$,slope of the tangent is $\frac{dy}{dx}=\frac{1}{2}$

equation of tangent at $(1,-2)$ is $(y+2)=\frac{1}{2}(x-1)$

$=>x=2y+5--------------\left(i\right)$

let $(x,y)$ be a point on the curve where the tangent to the first circle intersect/touches the second circle.

therefore,

$=>x^2+y^2+8x+6y+20=0$

put the value of $x$

$=>(5+2y{)}^2+y^2-8(5+2y)+6y+20=0$

$=>5y^2+10y+5=0$

$=>y^2+2y+1=0$

Three conditions of the discriminant($D$) of the above quadratic equation exists -

• If $D<0$, the tangent to the first circle neither intersects nor touches the second circle.
• If $D=0$, the tangent to the first circle touches the second circle.
• If $D>0$, the tangent to the first circle intersects the second circle.

In the above quadratic equation, $D=0$, hence, the tangent to the first circle at $(1,-2)$ touches the second circle