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  • The co-ordinate of the point from which the length of tangent to the following three circle be equal ,is <img alt="\mathrm{\begin{aligned} & 3 x^2+3 y^2+4 x-6 y-1=0 \\ & 2 x^2+2 y

The co-ordinate of the point from which the length of tangent to the following three circle be equal ,is

\mathrm{\begin{aligned} & 3 x^2+3 y^2+4 x-6 y-1=0 \\ & 2 x^2+2 y^2-3 x-2 y-4=0 \\ & 2 x^2+2 y^2-x+y-1=0 \end{aligned}}

Option: 1

\left(\frac{-16}{21}, \frac{-31}{63}\right)


Option: 2

\left(\frac{-8}{21}, \frac{-31}{63}\right)


Option: 3

(-2,-3)


Option: 4

(-4,6)


Answers (1)

best_answer

Here we have to find the radical centre of the three circles. First reduce them to standard form in which coefficient  of \mathrm{x^{2}} and \mathrm{y^{2}} be each unity. Subtracting in pairs the three radical axes are

\mathrm{\begin{aligned} & \frac{17}{6} x-y+\frac{5}{3}=0 ;-x-\frac{3}{2} y-\frac{3}{2}=0 \\ & -\frac{11}{6} x+\frac{5}{2} y-\frac{1}{6}=0 \end{aligned}}

Solving any two , we get the point \mathrm{\left(\frac{-16}{21}, \frac{-31}{63}\right)}which
satisfies the third also. This point is called the radical
centre and by definition the length of the tangents from it
to three circles are equal. 

Posted by

Ritika Kankaria

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