2x - 3y + 1 = 0 and 4x - 5y - 1 = 0 are the e | vedclass

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(B) The equations of the diameters of the circle are:$2x - 3y + 1 = 0$ ...$(i)$$4x - 5y - 1 = 0$ ...(ii)Solving equations $(i)$ and (ii),we find the c

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(B) The equations of the diameters of the circle are:$2x - 3y + 1 = 0$ ...$(i)$$4x - 5y - 1 = 0$ ...(ii)Solving equations $(i)$ and (ii),we find the centre of the circle $C = (-g, -f) = (3, 4)$,which implies $g = -3$ and $f = -4$.The equation of the circle is $x^2 + y^2 - 6x - 8y - 11 = 0$.The radius $r$ of the circle is $CQ = \sqrt{g^2 + f^2 - c} = \sqrt{(-3)^2 + (-4)^2 - (-11)} = \sqrt{9 + 16 + 11} = \sqrt{36} = 6$.The distance $CP$ between the centre $C(3, 4)$ and point $P(-2, -2)$ is $CP = \sqrt{(3 - (-2))^2 + (4 - (-2))^2} = \sqrt{5^2 + 6^2} = \sqrt{25 + 36} = \sqrt{61}$.In the right-angled triangle $	riangle CQP$,the length of the tangent $PQ$ is $PQ = \sqrt{CP^2 - CQ^2} = \sqrt{61 - 36} = \sqrt{25} = 5$.The area of the quadrilateral $PQCR$ is the sum of the areas of $	riangle CQP$ and $	riangle CRP$.Area of $PQCR = 2 	imes 	ext{Area}(	riangle CQP) = 2 	imes (\frac{1}{2} 	imes CQ 	imes PQ) = 6 	imes 5 = 30$ square units.

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