The intercept on the line y = x by the circle | vedclass

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Question

(a) Equation of any circle passing through the point of intersection of ${x^2} + {y^2} - 2x = 0$ and $y = x$ is

${x^2} + {y^2} - 2x + \lambda (y -

Vedclass · Accepted Answer

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

Vedclass · Answer

(a) Equation of any circle passing through the point of intersection of ${x^2} + {y^2} - 2x = 0$ and $y = x$ is

${x^2} + {y^2} - 2x + \lambda (y - x) = 0$

or ${x^2} + {y^2} - (2 + \lambda )x + \lambda y = 0$

Its center is $\left( {\frac{{2 + \lambda }}{2},\;\frac{{ - \lambda }}{2}} \right)$.

For $AB$ to be the diameter of the required circle, the centre must lie on $AB$

$i.e.$, $\frac{{2 + \lambda }}{2} = - \frac{\lambda }{2} $

$\Rightarrow \lambda = - 1$.

Thus the required equation of the circle is

${x^2} + {y^2} - 2x - y + x = 0$ or ${x^2} + {y^2} - x - y = 0$.

The intercept on the line $y = x$ by the circle ${x^2} + {y^2} - 2x = 0$ is $AB$ . Equation of the circle with $AB$ as a diameter is

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