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

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Question

(A) The equation of a family of circles passing through the intersection of the circle $S = x^2 + y^2 - a^2 = 0$ and the line $L = x - y + 3 = 0$ is g

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The equation of a family of circles 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$$x^2 + y^2 + \lambda x - \lambda y + (3\lambda - a^2) = 0$The center of this circle is $\left( -\frac{\lambda}{2}, \frac{\lambda}{2} \right)$.Since $AB$ is the diameter,the center of the circle must lie on the line $y = x + 3$.$\frac{\lambda}{2} = -\frac{\lambda}{2} + 3$$\lambda = 3$Substituting $\lambda = 3$ into the equation,we get:$(x^2 + y^2 - a^2) + 3(x - y + 3) = 0$$x^2 + y^2 + 3x - 3y - a^2 + 9 = 0$

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

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