Two tangents are drawn from the point P(-1,1) to the circle x^2+y^2-2x-6y+6=0.

Question

Two tangents are drawn from the point $P(-1,1)$ to the circle $x^2+y^2-2x-6y+6=0$. If these tangents touch the circle at points $A$ and $B$, and if $D$ is a point on the circle such that the lengths of $AB$ and $AD$ are equal, then the area of $	riangle ABD$ is equal to

Accepted Answer

Centre $(1,3)$, $r^2=1+9-6=4\Rightarrow r=2$. $PC=\sqrt{(1+1)^2+(3-1)^2}=2\sqrt2$, tangent length $PA=\sqrt{PC^2-r^2}=\sqrt{8-4}=2$.

$$AB=\frac{2r\cdot PA}{PC}=\frac{2\cdot2\cdot2}{2\sqrt2}=2\sqrt2.$$

Since $AD=AB=2\sqrt2$, both chords subtend $90^\circ$ at the centre (chord $=2r\sin(\theta/2)$). $B,D$ lie on opposite sides of $A$, making $BD$ a diameter ($=4$); by Thales' theorem $\angle BAD=90^\circ$.

$$\text{Area}(\triangle ABD)=\frac12\cdot AB\cdot AD=\frac12(2\sqrt2)(2\sqrt2)=4.$$

Solutions · 1