If the circle $x^2+y^2-6x-12y+1=0$ cuts another circle $C$ orthogonally and the centre of the circle $C$ is $(-4, 2)$,then its radius is

  • A
    $\sqrt{21}$
  • B
    $5$
  • C
    $\frac{3}{4}$
  • D
    $\sqrt{15}$

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Find the equation of the circle which cuts orthogonally each of the three circles $x^2+y^2-2x+3y-7=0$,$x^2+y^2+5x-5y+9=0$,and $x^2+y^2+7x-9y+29=0$.

If the centre of a circle,which passes through the points of intersection of the circles $x^2 + y^2 - 6x + 2y + 4 = 0$ and $x^2 + y^2 + 2x - 4y - 6 = 0$,lies on the line $y = x$,then the equation of the circle is:

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The length of the diameter of the circle which cuts the following three circles orthogonally is:
$x^{2}+y^{2}-x-y-14=0$
$x^{2}+y^{2}+3x-5y-10=0$
$x^{2}+y^{2}-2x+3y-27=0$

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