Let the tangents at the points A(4,-11) and B(8,-5) on the circle x^2+y^2-3x+10y-15=0 inte…

Question

Let the tangents at the points $A(4,-11)$ and $B(8,-5)$ on the circle $x^2+y^2-3x+10y-15=0$ intersect at the point $C$. Then the radius of the circle whose centre is $C$ and the line joining $A$ and $B$ is its tangent, is equal to:

Accepted Answer

Tangent at $A(4,-11)$: $4x-11y-\tfrac32(x+4)+5(y-11)-15=0\Rightarrow 5x-12y-152=0.$
Tangent at $B(8,-5)$: $8x-5y-\tfrac32(x+8)+5(y-5)-15=0\Rightarrow 13x-104=0\Rightarrow x=8.$

Substituting into the first: $40-12y-152=0\Rightarrow y=-\dfrac{28}{3}$. So $C=\left(8,-\dfrac{28}{3}\right)$.

Line $AB$: slope $=\dfrac{-5+11}{8-4}=\dfrac32$, giving $3x-2y-34=0$.

$$r=\frac{|3(8)-2(-28/3)-34|}{\sqrt{13}}=\frac{|{-26/3}|}{\sqrt{13}}=\frac{2\sqrt{13}}{3}.$$

Solutions · 1