If the squares of the lengths of the tangents | vedclass

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(C) Let the point be $P(x_1, y_1)$. The lengths of the tangents from $P$ to the circles $x^{2} + y^{2} = a^{2}$,$x^{2} + y^{2} = b^{2}$,and $x^{2} + y

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$a^{2}, b^{2}, c^{2}$ are in arithmetic progression

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(C) Let the point be $P(x_1, y_1)$. The lengths of the tangents from $P$ to the circles $x^{2} + y^{2} = a^{2}$,$x^{2} + y^{2} = b^{2}$,and $x^{2} + y^{2} = c^{2}$ are given by:$PT_1 = \sqrt{x_1^2 + y_1^2 - a^2}$$PT_2 = \sqrt{x_1^2 + y_1^2 - b^2}$$PT_3 = \sqrt{x_1^2 + y_1^2 - c^2}$Given that $PT_1^2, PT_2^2, PT_3^2$ are in arithmetic progression $(AP)$,we have:$2PT_2^2 = PT_1^2 + PT_3^2$Substituting the values:$2(x_1^2 + y_1^2 - b^2) = (x_1^2 + y_1^2 - a^2) + (x_1^2 + y_1^2 - c^2)$$2x_1^2 + 2y_1^2 - 2b^2 = 2x_1^2 + 2y_1^2 - a^2 - c^2$$-2b^2 = -a^2 - c^2$$2b^2 = a^2 + c^2$This implies that $a^2, b^2, c^2$ are in arithmetic progression.

If the squares of the lengths of the tangents drawn from a point $P$ to the circles $x^{2} + y^{2} = a^2$,$x^2 + y^{2} = b^2$,and $x^{2} + y^{2} = c^{2}$ are in arithmetic progression,then:

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