Find the co-ordinates of the point from which the lengths of tangents to the following three circles be equal
$3 x^{2} + 3 y^{2} + 4 x - 6 y - 1 = 0$
$2 x^{2} + 2 y^{2} - 3 x - 2 y - 4 = 0$
$2 x^{2} + 2 y^{2} - x + y - 1 = 0$ .

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Solution

Here we have to find the radical centre of the three circles. First reduce them to standard form in which coefficients of $x^{2}$ and $y^{2}$ be each unity.
Subtracting in pairs the three radical axes are
$\frac{17}{6} x - y + \frac{5}{3} = 0; - x - \frac{3}{2} y - \frac{3}{2} = 0$
$- \frac{11}{6} x + \frac{5}{2} y - \frac{1}{6} = 0$ .
Solving any two, we get the point $(\frac{- 16}{21}, \frac{- 31}{63})$
which satisfies the third also. This point is called the radical centre and by definition the length of the tangents from it to the three circles are equal.

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Find the co-ordinates of the point from which the lengths of tangents to the following three circles be equal3x2+3y2+4x−6y−1=02x2+2y2−3x−2y−4=02x2+2y2−x+y−1=0.

Find the co-ordinates of the point from which the lengths of tangents to the following three circles be equal
$3 x^{2} + 3 y^{2} + 4 x - 6 y - 1 = 0$
$2 x^{2} + 2 y^{2} - 3 x - 2 y - 4 = 0$
$2 x^{2} + 2 y^{2} - x + y - 1 = 0$ .