Let PQ and RS be tangents at the endpoints of | vedclass

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(C) Let the circle be $x^2 + y^2 = r^2$. The diameter $PR$ lies on the $x$-axis with $P = (-r, 0)$ and $R = (r, 0)$.The tangent at $P(-r, 0)$ is $x =

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$\frac{2PQ \cdot RS}{PQ + RS}$

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(C) Let the circle be $x^2 + y^2 = r^2$. The diameter $PR$ lies on the $x$-axis with $P = (-r, 0)$ and $R = (r, 0)$.The tangent at $P(-r, 0)$ is $x = -r$ and the tangent at $R(r, 0)$ is $x = r$.Let $Q = (-r, y_1)$ and $S = (r, y_2)$.The line $PS$ passes through $(-r, 0)$ and $(r, y_2)$,so its equation is $y - 0 = \frac{y_2 - 0}{r - (-r)}(x + r) \implies y = \frac{y_2}{2r}(x + r)$.The line $RQ$ passes through $(r, 0)$ and $(-r, y_1)$,so its equation is $y - 0 = \frac{y_1 - 0}{-r - r}(x - r) \implies y = \frac{y_1}{-2r}(x - r)$.Intersection $X(x, y)$ satisfies $y^2 = \frac{y_1 y_2}{4r^2}(r^2 - x^2)$. Since $X$ is on the circle,$y^2 = r^2 - x^2$.Thus,$r^2 - x^2 = \frac{y_1 y_2}{4r^2}(r^2 - x^2) \implies y_1 y_2 = 4r^2$.Since $PQ = |y_1|$ and $RS = |y_2|$,we have $PQ \cdot RS = 4r^2$.The length of the chord through $X$ perpendicular to $PR$ is $2|y| = 2\sqrt{r^2 - x^2}$.Using the geometry of the tangents,the height $h$ of $X$ satisfies $\frac{1}{h} = \frac{1}{PQ} + \frac{1}{RS} = \frac{PQ + RS}{PQ \cdot RS}$.Thus,$h = \frac{PQ \cdot RS}{PQ + RS}$. The full chord length is $2h = \frac{2PQ \cdot RS}{PQ + RS}$.

Let $PQ$ and $RS$ be tangents at the endpoints of the diameter $PR$ of a circle of radius $r$. If $PS$ and $RQ$ intersect at a point $X$ on the circumference of the circle,then the length of the chord through $X$ perpendicular to the diameter $PR$ is:

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