A circle C touches the line x=2y at the point | vedclass

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(A) The equation of a circle $C$ touching the line $x-2y=0$ at $(2,1)$ is given by $(x-2)^{2}+(y-1)^{2}+\lambda(x-2y)=0$.Expanding this,we get $x^{2}+

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$7\sqrt{5}$

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(A) The equation of a circle $C$ touching the line $x-2y=0$ at $(2,1)$ is given by $(x-2)^{2}+(y-1)^{2}+\lambda(x-2y)=0$.Expanding this,we get $x^{2}+y^{2}+x(\lambda-4)+y(-2-2\lambda)+5=0$.The given circle $C_{1}$ is $x^{2}+y^{2}+2y-5=0$.The common chord $PQ$ is the radical axis $C-C_{1}=0$,which is $x(\lambda-4)+y(-2-2\lambda-2)+10=0$,or $x(\lambda-4)-y(2\lambda+4)+10=0$.Since $PQ$ is a diameter of $C_{1}$,it must pass through the center of $C_{1}$,which is $(0,-1)$.Substituting $(0,-1)$ into the equation of $PQ$: $0(\lambda-4)-(-1)(2\lambda+4)+10=0$ $\Rightarrow 2\lambda+4+10=0$ $\Rightarrow 2\lambda=-14$ $\Rightarrow \lambda=-7$.Substituting $\lambda=-7$ into the equation of $C$: $x^{2}+y^{2}-11x+12y+5=0$.The center of $C$ is $(\frac{11}{2}, -6)$ and the radius $r$ is $\sqrt{(\frac{11}{2})^{2}+(-6)^{2}-5} = \sqrt{\frac{121}{4}+36-5} = \sqrt{\frac{121+124}{4}} = \sqrt{\frac{245}{4}} = \frac{7\sqrt{5}}{2}$.Thus,the diameter of $C$ is $2r = 7\sqrt{5}$.

$A$ circle $C$ touches the line $x=2y$ at the point $(2,1)$ and intersects the circle $C_{1}: x^{2}+y^{2}+2y-5=0$ at two points $P$ and $Q$ such that $PQ$ is a diameter of $C_{1}$. Then the diameter of $C$ is:

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