P is a point (a, b) in the first quadrant. If | vedclass

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(C) Let the equation of the two circles be $(x-r)^2 + (y-r)^2 = r^2$ (since they touch both axes,the center is $(r, r)$ and the radius is $r$).This si

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$a^2 - 4ab + b^2 = 0$

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(C) Let the equation of the two circles be $(x-r)^2 + (y-r)^2 = r^2$ (since they touch both axes,the center is $(r, r)$ and the radius is $r$).This simplifies to $x^2 + y^2 - 2rx - 2ry + r^2 = 0$.Let the radii of the two circles be $r_1$ and $r_2$. The condition for two circles $x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$ and $x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$ to cut orthogonally is $2g_1g_2 + 2f_1f_2 = c_1 + c_2$.Here,$g_1 = g_2 = -r_1, -r_2$ and $f_1 = f_2 = -r_1, -r_2$ and $c_1 = r_1^2, c_2 = r_2^2$.So,$2(-r_1)(-r_2) + 2(-r_1)(-r_2) = r_1^2 + r_2^2$,which gives $4r_1r_2 = r_1^2 + r_2^2$.Since the circles pass through $(a, b)$,we have $a^2 + b^2 - 2r(a+b) + r^2 = 0$,or $r^2 - 2r(a+b) + (a^2+b^2) = 0$.For this quadratic in $r$,the sum of roots $r_1 + r_2 = 2(a+b)$ and the product of roots $r_1r_2 = a^2 + b^2$.We know $r_1^2 + r_2^2 = (r_1 + r_2)^2 - 2r_1r_2$.Substituting into the orthogonality condition: $4r_1r_2 = (r_1 + r_2)^2 - 2r_1r_2$,so $6r_1r_2 = (r_1 + r_2)^2$.Substituting the values: $6(a^2 + b^2) = [2(a+b)]^2 = 4(a^2 + b^2 + 2ab)$.$6a^2 + 6b^2 = 4a^2 + 4b^2 + 8ab$.$2a^2 - 8ab + 2b^2 = 0$,which simplifies to $a^2 - 4ab + b^2 = 0$.

$P$ is a point $(a, b)$ in the first quadrant. If the two circles which pass through $P$ and touch both the coordinate axes cut at right angles,then:

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