Let O be the origin and OP and OQ be the tang | vedclass

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(B) The equation of the circle is $x^2+y^2-6x+4y+8=0$. The center of the circle is $C(3, -2)$.Since $OP$ and $OQ$ are tangents from the origin $O(0, 0

Vedclass · Accepted Answer

$\frac{5}{2}$

Vedclass · Answer

(B) The equation of the circle is $x^2+y^2-6x+4y+8=0$. The center of the circle is $C(3, -2)$.Since $OP$ and $OQ$ are tangents from the origin $O(0, 0)$ to the circle,the angle $\angle OPO = 90^\circ$ and $\angle OQO = 90^\circ$.Thus,$OP$ and $OQ$ subtend a right angle at the origin $O$ and at the center $C(3, -2)$.The circle with diameter $OC$ is the circumcircle of $	riangle OPQ$.The equation of the circle with diameter $OC$ where $O(0, 0)$ and $C(3, -2)$ is:$(x-0)(x-3) + (y-0)(y+2) = 0$$x^2 - 3x + y^2 + 2y = 0$$x^2 + y^2 - 3x + 2y = 0$Given that this circle passes through $(\alpha, \frac{1}{2})$,we substitute these coordinates into the equation:$\alpha^2 + (\frac{1}{2})^2 - 3\alpha + 2(\frac{1}{2}) = 0$$\alpha^2 + \frac{1}{4} - 3\alpha + 1 = 0$$\alpha^2 - 3\alpha + \frac{5}{4} = 0$$4\alpha^2 - 12\alpha + 5 = 0$$4\alpha^2 - 10\alpha - 2\alpha + 5 = 0$$2\alpha(2\alpha - 5) - 1(2\alpha - 5) = 0$$(2\alpha - 1)(2\alpha - 5) = 0$Therefore,$\alpha = \frac{1}{2}$ or $\alpha = \frac{5}{2}$.Comparing with the options,the correct value is $\frac{5}{2}$.

Let $O$ be the origin and $OP$ and $OQ$ be the tangents to the circle $x^2+y^2-6x+4y+8=0$ at the points $P$ and $Q$ on it. If the circumcircle of the triangle $OPQ$ passes through the point $(\alpha, \frac{1}{2})$,then a value of $\alpha$ is

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