A circle passing through the point P(alpha, b | vedclass

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(D) Let the equation of the circle be $(x-a)^2 + (y-a)^2 = a^2$,where $a$ is the radius of the circle.Since the circle passes through $P(\alpha, \beta

Vedclass · Accepted Answer

$121$

Vedclass · Answer

(D) Let the equation of the circle be $(x-a)^2 + (y-a)^2 = a^2$,where $a$ is the radius of the circle.Since the circle passes through $P(\alpha, \beta)$,we have $(\alpha-a)^2 + (\beta-a)^2 = a^2$.Expanding this,we get $\alpha^2 - 2\alpha a + a^2 + \beta^2 - 2\beta a + a^2 = a^2$,which simplifies to $\alpha^2 + \beta^2 - 2a(\alpha + \beta) + a^2 = 0$.The points of contact with the axes are $A(a, 0)$ and $B(0, a)$. The equation of the line $AB$ is $x + y = a$,or $x + y - a = 0$.The length of the perpendicular $PQ$ from $P(\alpha, \beta)$ to the line $x + y - a = 0$ is given by $PQ = \frac{|\alpha + \beta - a|}{\sqrt{1^2 + 1^2}} = \frac{|\alpha + \beta - a|}{\sqrt{2}}$.Given $PQ = 11$,we have $\frac{|\alpha + \beta - a|}{\sqrt{2}} = 11$,so $|\alpha + \beta - a| = 11\sqrt{2}$.Squaring both sides,$(\alpha + \beta - a)^2 = 242$.Expanding this,$\alpha^2 + \beta^2 + a^2 + 2\alpha\beta - 2a(\alpha + \beta) = 242$.From the circle equation,we know $\alpha^2 + \beta^2 - 2a(\alpha + \beta) = -a^2$.Substituting this into the squared equation: $-a^2 + a^2 + 2\alpha\beta = 242$.Thus,$2\alpha\beta = 242$,which gives $\alpha\beta = 121$.

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