Let B be the centre of the circle x^{2}+y^{2} | vedclass

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(A) The equation of the circle is $x^{2}+y^{2}-2x+4y+1=0$. Comparing with $x^{2}+y^{2}+2gx+2fy+c=0$,we get $g=-1, f=2, c=1$. The centre $B$ is $(-g, -

Vedclass · Accepted Answer

$18$

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(A) The equation of the circle is $x^{2}+y^{2}-2x+4y+1=0$. Comparing with $x^{2}+y^{2}+2gx+2fy+c=0$,we get $g=-1, f=2, c=1$. The centre $B$ is $(-g, -f) = (1, -2)$ and the radius $r = \sqrt{g^{2}+f^{2}-c} = \sqrt{1+4-1} = 2$.Let $AB$ intersect $PQ$ at $R$. Since $AB$ is the perpendicular bisector of $PQ$,$AR \perp PQ$ and $BR \perp PQ$.In $	riangle ABP$,$\angle APB = 90^{\circ}$ (tangent is perpendicular to radius).$AB = \sqrt{(3-1)^{2} + (1-(-2))^{2}} = \sqrt{2^{2} + 3^{2}} = \sqrt{4+9} = \sqrt{13}$.In right $	riangle ABP$,$AP = \sqrt{AB^{2} - BP^{2}} = \sqrt{13 - 2^{2}} = \sqrt{13-4} = 3$.In $	riangle ABP$,$AR$ is the altitude to the hypotenuse $BP$ is not correct,rather $PR$ is the altitude to $AB$. Using area of $	riangle ABP$,$AR 	imes BP = AP 	imes PR$ is not the right approach. Instead,$PR = \frac{AP 	imes BP}{AB} = \frac{3 	imes 2}{\sqrt{13}} = \frac{6}{\sqrt{13}}$.$AR = \sqrt{AP^{2} - PR^{2}} = \sqrt{9 - \frac{36}{13}} = \sqrt{\frac{117-36}{13}} = \sqrt{\frac{81}{13}} = \frac{9}{\sqrt{13}}$.$BR = \sqrt{BP^{2} - PR^{2}} = \sqrt{4 - \frac{36}{13}} = \sqrt{\frac{52-36}{13}} = \sqrt{\frac{16}{13}} = \frac{4}{\sqrt{13}}$.Area $	riangle APQ = \frac{1}{2} 	imes PQ 	imes AR = PR 	imes AR = \frac{6}{\sqrt{13}} 	imes \frac{9}{\sqrt{13}} = \frac{54}{13}$.Area $	riangle BPQ = \frac{1}{2} 	imes PQ 	imes BR = PR 	imes BR = \frac{6}{\sqrt{13}} 	imes \frac{4}{\sqrt{13}} = \frac{24}{13}$.Ratio $\frac{	ext{Area } 	riangle APQ}{	ext{Area } 	riangle BPQ} = \frac{54/13}{24/13} = \frac{54}{24} = \frac{9}{4}$.Thus,$8 \left(\frac{	ext{Area } 	riangle APQ}{	ext{Area } 	riangle BPQ}ight) = 8 	imes \frac{9}{4} = 18$.

Let $B$ be the centre of the circle $x^{2}+y^{2}-2x+4y+1=0$. Let the tangents at two points $P$ and $Q$ on the circle intersect at the point $A(3,1)$. Then $8 \left(\frac{\text{Area } \triangle APQ}{\text{Area } \triangle BPQ}\right)$ is equal to:

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