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(B) The given circle is $x^2+y^2-4x-2y-4=0$. The center of this circle is $(2,1)$. Since the diameter passes through the origin $(0,0)$ and the center

Vedclass · Accepted Answer

$3x^2+3y^2-19x+8y-12=0$

Vedclass · Answer

(B) The given circle is $x^2+y^2-4x-2y-4=0$. The center of this circle is $(2,1)$. Since the diameter passes through the origin $(0,0)$ and the center $(2,1)$,the equation of the diameter is $y = \frac{1}{2}x$,which simplifies to $x-2y=0$. The family of circles passing through the intersection of the circle $S=0$ and the line $L=0$ is given by $S+\lambda L=0$. Thus,the equation is $x^2+y^2-4x-2y-4+\lambda(x-2y)=0$. Since this circle passes through the point $(1,2)$,we substitute $x=1$ and $y=2$: $1^2+2^2-4(1)-2(2)-4+\lambda(1-2(2))=0$ $1+4-4-4-4+\lambda(1-4)=0$ $-7-3\lambda=0 \implies \lambda = -\frac{7}{3}$. Substituting $\lambda$ back into the equation: $x^2+y^2-4x-2y-4-\frac{7}{3}(x-2y)=0$ $3(x^2+y^2)-12x-6y-12-7x+14y=0$ $3x^2+3y^2-19x+8y-12=0$.

If the origin lies on a diameter of the circle $x^2+y^2-4x-2y-4=0$,then the equation of the circle passing through the end points of that diameter and the point $(1,2)$ is

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