$(a, b)$ is the midpoint of the chord AB of the circle $x^{2} + y^{2} = r^{2}$ . The tangents at A,B meet at C, then area of $∆ A B C =$

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\frac{{(a^{2} + b^{2} + r^{2})}^{3 / 2}}{\sqrt{a^{2} + b^{2}}}$

$\frac{{(r^{2} - a^{2} - b^{2})}^{3 / 2}}{\sqrt{a^{2} + b^{2}}}$

$\frac{{(a^{2} - b^{2} - r^{2})}^{3 / 2}}{\sqrt{a^{2} + b^{2}}}$

none

answer is B.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Equation of the chord AB with (a, b) as midpoint is $a x + b y - (a^{2} + b^{2}) = 0$

d = Distance from $C (0, 0) to \overset{\leftrightarrow}{A B} = \frac{|- (a^{2} + b^{2})|}{\sqrt{a^{2} + b^{2}}} = \sqrt{a^{2} + b^{2}}$

AB = Length of the chord $= 2 \sqrt{r^{2} - d^{2}} = 2 \sqrt{r^{2} - a^{2} - b^{2}}$

Pole of $a x + b y - (a^{2} + b^{2}) = 0$ with respect to $x^{2} + y^{2} = r^{2} is C (\frac{- a r^{2}}{a^{2} + b^{2}}, \frac{b r^{2}}{a^{2} + b^{2}})$

h = Distance from C to $a x + b y - (a^{2} + b^{2}) = 0$

$= \frac{|\frac{a^{2} r^{2}}{a^{2} + b^{2}} + \frac{b^{2} r^{2}}{a^{2} + b^{2}} - (a^{2} + b^{2})|}{\sqrt{a^{2} + b^{2}}} = \frac{|\frac{r^{2} (a^{2} + b^{2})}{a^{2} + b^{2}} - a^{2} - b^{2}|}{\sqrt{a^{2} + b^{2}}} = \frac{r^{2} - a^{2} - b^{2}}{\sqrt{a^{2} + b^{2}}}$

Area of $Δ A B C = \frac{1}{2} \times h \times A B = \frac{1}{2} \times \frac{r^{2} - a^{2} - b^{2}}{\sqrt{a^{2} + b^{2}}} \times 2 \sqrt{r^{2} - a^{2} - b^{2}} = \frac{{(r^{2} - a^{2} - b^{2})}^{3 / 2}}{\sqrt{a^{2} + b^{2}}}$