If the line y = x + 3 intersects the circle x | vedclass

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Question

(A) The equation of any circle passing through the intersection of the circle $S: x^2 + y^2 - a^2 = 0$ and the line $L: x - y + 3 = 0$ is given by $S

Vedclass · Accepted Answer

$x^2 + y^2 + 3x - 3y - a^2 + 9 = 0$

Vedclass · Answer

(A) The equation of any circle passing through the intersection of the circle $S: x^2 + y^2 - a^2 = 0$ and the line $L: x - y + 3 = 0$ is given by $S + \lambda L = 0$.$x^2 + y^2 - a^2 + \lambda(x - y + 3) = 0$$x^2 + y^2 + \lambda x - \lambda y + (3\lambda - a^2) = 0$The center of this circle is $\left(-\frac{\lambda}{2}, \frac{\lambda}{2}\right)$.Since $AB$ is the diameter,the center must lie on the line $y = x + 3$.Substituting the center coordinates into the line equation: $\frac{\lambda}{2} = -\frac{\lambda}{2} + 3$.$\lambda = 3$.Substituting $\lambda = 3$ into the circle equation: $x^2 + y^2 + 3x - 3y + (3(3) - a^2) = 0$.$x^2 + y^2 + 3x - 3y + 9 - a^2 = 0$.

If the line $y = x + 3$ intersects the circle $x^2 + y^2 = a^2$ at points $A$ and $B$,then the equation of the circle with $AB$ as its diameter is:

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