Question

Tangents are drawn from $(4,4)$ to the circle $x^2+y^2-2x-2y-7=0$ to meet the circle at $A$ and $B$ . The length of the chord $AB$ is :

• A. $2\sqrt{3}$
• B. $3\sqrt{2}$
• C. $2\sqrt{6}$
• D. $6\sqrt{2}$

Answer 1

The correct option is A $2\sqrt{3}$

$x^2+y^2-2x-2y-7=0$

$(x-1{)}^2,(y-1{)}^2=3^2$

Centre $(1,1)$ radius $=3units$

$A{P}^2=(OP{)}^2-(OA{)}^2=(\sqrt{(4-1{)}^2+(4-1{)}^2}{)}^2-3^2$

$A{P}^2=3^2$

$AP=3units$

Similarly $B{P}^2-O{P}^2-O{B}^2=(\sqrt{(4-1{)}^2+(4-1{)}^2}{)}^2-3^2$

$B{P}^2=3^2$

$BP=3units$

Consider triangle $OAP$

Area of $\left(△OAP\right)=\frac{1}{2}\times base\times height$

$=\frac{1}{2}\times OP\times AQ=\frac{1}{2}\left(OA\right)\times \left(AP\right)$

$\Rightarrow \left(OP\right)\left(AQ\right)=\left(OA\right)\left(AP\right)$

$\Rightarrow \left(3\sqrt{2}\right)\left(AQ\right)=3\left(3\right)$

$\Rightarrow \left(AQ\right)=\frac{3}{\sqrt{2}}unit$

But we know $AQ=\frac{1}{2}AB$

(perpendicular from centre bisect chord)

$\Rightarrow \frac{3}{\sqrt{2}}=\frac{1}{2}AB$

$\Rightarrow AB=3\sqrt{2}units$