Question

Jee level question
Chapter:- Coordinate geometry ​

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that two circles intersect each other at A and B.

↝ So that, Length of common chord be AB.

Further,

Let assume that centre of circles of radius 5 cm and 12 cm be P and Q respectively.

Let further assume that line segment joining centres P and Q intersect AB at R respectively.

We know,

Line segment joining the centres of two intersecting circles is perpendicular bisector of the common chord.

It means, PQ is perpendicular bisector of AB.

Let assume that, AR = x and PR = y and ∠APQ = p

Now, In right angle triangle APQ,

$\rm :\longmapsto\:tanp = \dfrac{AQ}{AP}$

$\rm :\longmapsto\:tanp = \dfrac{12}{5}$

We know,

$\rm :\longmapsto\:cotp = \dfrac{1}{tanp}$

$\bf\implies \:cotp = \dfrac{5}{12}$

Further, we know that

$\rm :\longmapsto\: {cosec}^{2}p - {cot}^{2}p = 1$

$\rm :\longmapsto\: {cosec}^{2}p - \dfrac{25}{144} = 1$

$\rm :\longmapsto\: {cosec}^{2}p = \dfrac{25}{144} + 1$

$\rm :\longmapsto\: {cosec}^{2}p = \dfrac{25 + 144}{144}$

$\rm :\longmapsto\: {cosec}^{2}p = \dfrac{169}{144}$

As triangle APQ is right angled triangle, so ∠APQ < 90°.

So, cosecp > 0

Hence,

$\rm :\longmapsto\: {cosec}p = \dfrac{13}{12}$

$\bf\implies \:sinp = \dfrac{12}{13}$

Now, in right angle triangle APR

$\rm :\longmapsto\:sinp = \dfrac{AR}{AP}$

$\rm :\longmapsto\:\dfrac{12}{13} = \dfrac{x}{5}$

$\bf\implies \:x = \dfrac{60}{13}$

Hence,

$\boxed{ \bf{ \: Length \: of \: common \: chord, \: AB = 2x = \dfrac{120}{13}}}$

Thus,

$\: \: \: \: \: \: \: \: \: \: \: \: \: \underbrace{ \boxed{ \bf \: Option \: (b) \: is \: correct \: \: }}$