If y = 2x is a chord of the circle {x^2} + {y | vedclass

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Question

(d) Family of circles passing from points of intersection of circle and chord

are ${x^2} + {y^2} - 10x + \lambda (y - 2x) = 0$.

Its centre is $\

Vedclass · Accepted Answer

${x^2} + {y^2} - 2x - 4y = 0$

Vedclass · Answer

(d) Family of circles passing from points of intersection of circle and chord

are ${x^2} + {y^2} - 10x + \lambda (y - 2x) = 0$.

Its centre is $\left\{ {(5 + \lambda ),\; - \frac{\lambda }{2}} \right\}$. It lies on $y - 2x = 0$.

i.e. $ - \frac{\lambda }{2} - 10 - 2\lambda = 0 \Rightarrow - 5\lambda = 20 \Rightarrow \lambda = - 4$

Hence equation of circle is ${x^2} + {y^2} - 2x - 4y = 0$.

Trick : Since $y = 2x$ is the line passing through origin and through $I$ and $III$ quadrant. Also centre of the given circle lies on positive x-axis, therefore, the centre of required circle will be in I quadrant and must satisfy the given line $y = 2x$ and

its equation is given by ${x^2} + {y^2} - 2x - 4y = 0$.

If $y = 2x$ is a chord of the circle ${x^2} + {y^2} - 10x = 0$, then the equation of the circle of which this chord is a diameter, is

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