Let AB be a chord of length 12 of the circle | vedclass

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Question

(C) The radius of the circle $r = \sqrt{\frac{169}{4}} = \frac{13}{2}$.Let $C$ be the center of the circle and $M$ be the midpoint of the chord $AB$.

Vedclass · Accepted Answer

$72$

Vedclass · Answer

(C) The radius of the circle $r = \sqrt{\frac{169}{4}} = \frac{13}{2}$.Let $C$ be the center of the circle and $M$ be the midpoint of the chord $AB$. Since $AB = 12$,$AM = 6$.In the right-angled triangle $	riangle AMC$,$AC = \frac{13}{2}$ and $AM = 6$.Using the Pythagorean theorem,$CM = \sqrt{AC^2 - AM^2} = \sqrt{(\frac{13}{2})^2 - 6^2} = \sqrt{\frac{169}{4} - 36} = \sqrt{\frac{169-144}{4}} = \sqrt{\frac{25}{4}} = \frac{5}{2}$.Let $\angle ACM = 	heta$. Then $\sin 	heta = \frac{AM}{AC} = \frac{6}{13/2} = \frac{12}{13}$ and $\cos 	heta = \frac{CM}{AC} = \frac{5/2}{13/2} = \frac{5}{13}$.In $	riangle PAC$,$\angle PAC = 90^\circ$ because $PA$ is a tangent. Thus,$	riangle PAC$ is a right-angled triangle.In $	riangle PAC$,$AM$ is the altitude to the hypotenuse $PC$. By property of right triangles,$AM^2 = PM \cdot MC$.$6^2 = PM \cdot \frac{5}{2} \implies 36 = PM \cdot \frac{5}{2} \implies PM = \frac{72}{5}$.The distance of point $P$ from chord $AB$ is $PM$.Therefore,$5(PM) = 5 \cdot \frac{72}{5} = 72$.

Let $AB$ be a chord of length $12$ of the circle $(x-2)^{2}+(y+1)^{2}=\frac{169}{4}$. If tangents drawn to the circle at points $A$ and $B$ intersect at the point $P$,then five times the distance of point $P$ from chord $AB$ is equal to $.......$

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