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

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(A) The equation of any circle passing through the intersection of the circle $x^2 + y^2 - 2x = 0$ and the line $y - x = 0$ is given by the family of

Vedclass · Accepted Answer

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

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(A) The equation of any circle passing through the intersection of the circle $x^2 + y^2 - 2x = 0$ and the line $y - x = 0$ is given by the family of circles equation: $x^2 + y^2 - 2x + \lambda(y - x) = 0$ $x^2 + y^2 - (2 + \lambda)x + \lambda y = 0$ The center of this circle is $\left( \frac{2 + \lambda}{2}, -\frac{\lambda}{2} \right)$. Since $AB$ is the diameter,the center of the circle must lie on the line $y = x$. Substituting the center coordinates into $y = x$: $-\frac{\lambda}{2} = \frac{2 + \lambda}{2} -\lambda = 2 + \lambda 2\lambda = -2 \lambda = -1$ Substituting $\lambda = -1$ back into the family equation: $x^2 + y^2 - (2 - 1)x - 1y = 0$ $x^2 + y^2 - x - y = 0$.

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

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