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10-1.Circle and System of Circles
hard
The equation of the circle through the point of intersection of the circles ${x^2} + {y^2} - 8x - 2y + 7 = 0$, ${x^2} + {y^2} - 4x + 10y + 8 = 0$ and $(3, -3)$ is
A
$23{x^2} + 23{y^2} - 156x + 38y + 168 = 0$
B
$23{x^2} + 23{y^2} + 156x + 38y + 168 = 0$
C
${x^2} + {y^2} + 156x + 38y + 168 = 0$
D
None of these
Solution
(a) Equation of circle is
$({x^2} + {y^2} – 8x – 2y + 7) + \lambda ({x^2} + {y^2} – 4x + 10y + 8) = 0$
Also point $(3,\; – 3)$ lies on the above equation.
$ \Rightarrow \lambda = \frac{7}{{16}}$
Hence required equation is
$23{x^2} + 23{y^2} – 156x + 38y + 168 = 0$.
Standard 11
Mathematics
