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  • Let the tangents at the points <img alt="\mathrm{B(4, -11) \: and \: B(8, -5)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7BB%284%2C%20-11%29%20%5C%3A%20and%20%5C%3A%20B%288%2C

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

Option: 1

2\sqrt{13}


Option: 2

\sqrt{13}


Option: 3

\frac{3\sqrt{3}}{4}


Option: 4

\frac{2\sqrt{13}}{3}


Answers (1)

best_answer

Equation of line \mathrm{AB} is

\mathrm{y+5=\left ( \frac{-5+11}{8-4} \right )\left ( x-8 \right )}

\mathrm{y+5=\frac{3}{2}\left ( x-8 \right )}

\mathrm{3 x-2 y-34=0}.................(i)
Let \mathrm{C}$ be $(\mathrm{h}, \mathrm{k}) then equation of \mathrm{AB}
\begin{aligned} & \mathrm{hx}+\mathrm{ky}-\frac{3}{2}(\mathrm{x}+\mathrm{h})+5(\mathrm{y}+\mathrm{k})-15=0 \\ & \mathrm{x}\left(\mathrm{h}-\frac{3}{2}\right)+\mathrm{y}(\mathrm{k}+5)-\frac{3}{2} \mathrm{~h}+5 \mathrm{k}-15=0 \end{aligned}..................(ii)
Now, by comparing (i) and (ii)

\frac{h-\frac{3}{2}}{3}=\frac{k+5}{-2}=\frac{-\frac{3}{2} h+5 k-15}{-34}
after solving centre \mathrm{C} is
(\mathrm{h}, \mathrm{k})=\left(8, \frac{-28}{3}\right)
and radius of circle is
\begin{aligned} r & =\left|\frac{3(8)-2\left(\frac{-28}{3}\right)-34}{\sqrt{9+4}}\right|=\left|\frac{24+2 \frac{56}{3}-34}{\sqrt{13}}\right| \\ r & =\left|\frac{26}{3 \sqrt{13}}\right|=\frac{2 \sqrt{13}}{3} \end{aligned}

Posted by

jitender.kumar

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