Question

$AB$ is a chord of the circle $x^2+y^2=9$. The tangent at $A$ and $B$ intersect at $C$. lf $(1,2)$ is the midpoint of $AB$, then the area of $\Delta ABC$ is (in square units)

• A. $9$
• B. $\frac{8}{\sqrt{5}}$
• C. $\frac{9}{\sqrt{5}}$
• D. $\frac{7}{\sqrt{3}}$

Answer 1

The correct option is B $\frac{8}{\sqrt{5}}$

Let, $M(1,2)$is a mid-point of AB, than O,M and C are on same line
$OM=\sqrt{5}$ and radius of circle$=3$
By definition of pole and polor,
C is a pole of chord AB
$\therefore OM.OC=r^2=9$
$OC=\frac{9}{\sqrt{5}}$
$\therefore CM=OC-OM=\frac{9}{\sqrt{5}}\sqrt{5}=\frac{4}{\sqrt{5}}$
In $\Delta OAM,$
$A{M}^2+O{M}^2=O{A}^2=9$
$A{M}^2=9-5=4$
$AM=2$

so, length of chord $AB=2AM=4$

$\therefore$ Area of $\Delta ABC=\frac{1}{2}AB\times CM$

$=\frac{1}{2}\times 4\times \frac{4}{\sqrt{5}}=\frac{8}{\sqrt{5}}$ sq units