Tangents drawn from the origin to the circle | vedclass

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(C) The equation of the circle is $x^2 + y^2 - 2ax - 2by + b^2 = 0$.Comparing this with the general equation $x^2 + y^2 + 2gx + 2fy + c = 0$,we get th

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$a^2 = b^2$

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(C) The equation of the circle is $x^2 + y^2 - 2ax - 2by + b^2 = 0$.Comparing this with the general equation $x^2 + y^2 + 2gx + 2fy + c = 0$,we get the center $(g, f) = (a, b)$ and radius $r = \sqrt{a^2 + b^2 - b^2} = \sqrt{a^2} = |a|$.For tangents drawn from the origin $(0, 0)$ to be perpendicular,the distance from the center $(a, b)$ to the origin must be equal to $\sqrt{2} 	imes r$.Distance from $(0, 0)$ to $(a, b)$ is $\sqrt{a^2 + b^2}$.So,$\sqrt{a^2 + b^2} = \sqrt{2} 	imes |a|$.Squaring both sides,we get $a^2 + b^2 = 2a^2$,which simplifies to $b^2 = a^2$ or $a^2 = b^2$.

Tangents drawn from the origin to the circle $x^2 + y^2 - 2ax - 2by + b^2 = 0$ are perpendicular to each other,if

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