P, Q and R are the centres and r_1, r_2, r_3 | vedclass

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(B) For three co-axial circles with centres $P, Q, R$ and radii $r_1, r_2, r_3$,the radical axis is common.Let the centres be at $x_1, x_2, x_3$ on th

Vedclass · Accepted Answer

$-PQ \cdot QR \cdot RP$

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(B) For three co-axial circles with centres $P, Q, R$ and radii $r_1, r_2, r_3$,the radical axis is common.Let the centres be at $x_1, x_2, x_3$ on the $x$-axis.The power of a point on the radical axis is constant,so $x_i^2 - r_i^2 = k$ for $i = 1, 2, 3$.Thus,$r_i^2 = x_i^2 - k$.The expression $QR r_1^2 + RP r_2^2 + PQ r_3^2$ becomes $(x_3 - x_2)(x_1^2 - k) + (x_1 - x_3)(x_2^2 - k) + (x_2 - x_1)(x_3^2 - k)$.Since $(x_3 - x_2) + (x_1 - x_3) + (x_2 - x_1) = 0$,the terms involving $k$ cancel out.We are left with $(x_3 - x_2)x_1^2 + (x_1 - x_3)x_2^2 + (x_2 - x_1)x_3^2$.This is a cyclic expression equal to $-(x_1 - x_2)(x_2 - x_3)(x_3 - x_1)$.Given $PQ = |x_1 - x_2|$,$QR = |x_2 - x_3|$,and $RP = |x_3 - x_1|$,the result is $-PQ \cdot QR \cdot RP$.

$P, Q$ and $R$ are the centres and $r_1, r_2, r_3$ are the radii respectively of three co-axial circles. Then $QRr_1^2 + RP r_2^2 + PQ r_3^2$ is equal to

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