JEE MAIN - Mathematics (2023 - 29th January Morning Shift - No. 6)

Let the tangents at the points $$A(4,-11)$$ and $$B(8,-5)$$ on the circle $$x^{2}+y^{2}-3 x+10 y-15=0$$, intersect at the point $$C$$. Then the radius of the circle, whose centre is $$C$$ and the line joining $$A$$ and $$B$$ is its tangent, is equal to :
$$\frac{2\sqrt{13}}{3}$$
$$\frac{3\sqrt{3}}{4}$$
$$\sqrt{13}$$
$$2\sqrt{13}$$

Explanation

JEE Main 2023 (Online) 29th January Morning Shift Mathematics - Circle Question 40 English Explanation

Equation of AB :

$$y - ( - 5) = {{ - 5 + 11} \over {8 - 4}}(x - 8)$$

$$ \Rightarrow y + 5 = {3 \over 2}(x - 8)$$

$$ \Rightarrow 2y + 10 = 3x - 24$$

$$ \Rightarrow 3x - 2y - 34 = 0$$ .... (1)

Also AB is cord of contact. And we know equation of cord of contact to a circle is $$T = 0$$

$$ \Rightarrow xh + yk - {3 \over 2}(x + h) + 5(y + k) - 15 = 0$$

$$ \Rightarrow x\left( {h - {3 \over 2}} \right) + y(k + 5) + \left( { - {3 \over 2}h + 5k - 15} \right) = 0$$ .... (2)

Comparing equation (1) and (2), we get

$${{h - {3 \over 2}} \over 3} = {{k + 5} \over { - 2}} = {{ - {3 \over 2}h + 5k - 15} \over { - 34}}$$

$$\therefore$$ $$ - 2h + 3 = 3k + 15$$

$$ \Rightarrow 3k + 2h = - 12$$ ..... (3)

and

$$17(k + 5) = - {3 \over 2}h + 5k - 15$$

$$ \Rightarrow 12k + {3 \over 2}h - 100$$

$$ \Rightarrow 3h + 24k = - 200$$ ..... (4)

Solving (3) and (4), we get

$$h = 8$$ and $$k = - {{28} \over 3}$$

$$\therefore$$ Point C is $$\left( {8, - {{28} \over 3}} \right)$$

Now radius of the circle whose centre is at C and tangent is AB is

$$ = \left| {{{3(8) - 2\left( { - {{28} \over 3}} \right) - 34} \over {\sqrt {{3^2} + {2^2}} }}} \right|$$

$$ = \left| {{{26} \over {3\sqrt {13} }}} \right|$$

$$ = {{2\sqrt {13} } \over 3}$$

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