A triangle PQR is inscribed in the circle x^2 | vedclass

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(C) The circle is $x^2 + y^2 = 5^2$,with center $O(0, 0)$ and radius $r = 5$.Let $Q = (3, 4)$ and $R = (-4, 3)$.The slope of $OQ$ is $m_1 = \frac{4-0}

Vedclass · Accepted Answer

$\pi /4$

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(C) The circle is $x^2 + y^2 = 5^2$,with center $O(0, 0)$ and radius $r = 5$.Let $Q = (3, 4)$ and $R = (-4, 3)$.The slope of $OQ$ is $m_1 = \frac{4-0}{3-0} = \frac{4}{3}$.The slope of $OR$ is $m_2 = \frac{3-0}{-4-0} = -\frac{3}{4}$.Since $m_1 	imes m_2 = \frac{4}{3} 	imes (-\frac{3}{4}) = -1$,the lines $OQ$ and $OR$ are perpendicular.Thus,the central angle $\angle QOR = \frac{\pi}{2}$.By the circle theorem,the angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.Therefore,$\angle QPR = \frac{1}{2} \angle QOR = \frac{1}{2} 	imes \frac{\pi}{2} = \frac{\pi}{4}$.

$A$ triangle $PQR$ is inscribed in the circle $x^2 + y^2 = 25$. If the coordinates of $Q$ and $R$ are $(3, 4)$ and $(-4, 3)$ respectively,then $\angle QPR = \dots$

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