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Tangents are drawn to the circle \mathrm{x^2+y^2=10} at the points where it is met by the circle \mathrm{x^2+y^2+4 x-3 y+2=0}. The point of intersection of these tangents is
 

Option: 1

\mathrm{\left(\frac{5}{2}, \frac{-10}{3}\right)}

 


Option: 2

\left(\frac{5}{2}, \frac{10}{3}\right)
 


Option: 3

\left(-\frac{10}{3}, \frac{5}{2}\right)
 


Option: 4

\left(-\frac{10}{3}, \frac{5}{2}\right)


Answers (1)

best_answer

Equation of common chord is

\mathrm{ S-S^{\prime}=0 }

\mathrm{ 4 x-3 y+12=0}

Let the required point be \mathrm{ (h, k)}. Then its chord of contact w.r.t circle :

\mathrm{x^2+y^2=10 \: is \: \mathrm{hx}+\mathrm{ky}-10=0}

Comparing the two equation, we get

\mathrm{ \frac{\mathrm{h}}{4}=\frac{\mathrm{k}}{-3}=\frac{-10}{12} }

\mathrm{ \text { i.e. } h=-\frac{10}{3}, k=\frac{5}{2} }

\mathrm{ i.e. \left(-\frac{10}{3}, \frac{5}{2}\right) } is the required point of intersection.

Hence option 3 is correct.

Posted by

Ritika Harsh

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