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If the origin lies on a diameter of the circle $x^2+y^2-4 x-2 y-4=0$, then the equation of the circle passing through the end points of that diameter and the point $(1,2)$ is
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Verified Answer
The correct answer is:
$3 x^2+3 y^2-19 x+8 y-12=0$
Given circle
$x^2+y^2-4 x-2 y-4=0$
Centre $(2,1)$
Equation of diameter of circle passes through origin.
$x-2 y=0$
Equation of circle through the end point of diameter of circle and line $x-2 y=0$ is
$x^2+y^2-4 x-2 y-4+\lambda(x-2 y)=0$
Since, this is passes through $(1,2)$
$\because \quad 1+4-4-4-4+\lambda(-3)=0$
$\lambda=\frac{-7}{3}$
Required equation of circle is
$x^2+y^2-4 x-2 y-4-\frac{7}{2}(x-2 y)=0$
$3\left(x^2+y^2\right)-19 x+8 y-12=0$
$3 x^2+3 y^2-19 x+8 y-12=0$
$x^2+y^2-4 x-2 y-4=0$
Centre $(2,1)$
Equation of diameter of circle passes through origin.
$x-2 y=0$
Equation of circle through the end point of diameter of circle and line $x-2 y=0$ is
$x^2+y^2-4 x-2 y-4+\lambda(x-2 y)=0$
Since, this is passes through $(1,2)$
$\because \quad 1+4-4-4-4+\lambda(-3)=0$
$\lambda=\frac{-7}{3}$
Required equation of circle is
$x^2+y^2-4 x-2 y-4-\frac{7}{2}(x-2 y)=0$
$3\left(x^2+y^2\right)-19 x+8 y-12=0$
$3 x^2+3 y^2-19 x+8 y-12=0$
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