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VITEEE 2012 Paper

Section: Mathematics

Question : 92 of 120

Marks: +1, -0

Solution:

Center is the point of intersection of diagonals.

\Rightarrow 2x-3y=5 ...... (i)

\Rightarrow 3x-4y=7 ........(ii)

are diameter of circle so, point of intersection is centre of circle.from (1) & (2), x=1, y=−1

∴ area of circle is 154

= (\pi r)^2 = 154

r^2 = \frac{7}{22} \times 154

= \frac{7 \times 14}{2}

r^2 = 49

⇒ r = 7

∴ Equation of circle with centre (1,−1) and radius 7 units

\Rightarrow (x-1)^2+(y+1)^2=7^2

\Rightarrow x^{2}+1-2x+y^{2}+1+2y=4 9

\Rightarrow x^{2}+y^{2}-2x+2y=4 9 - 2

\Rightarrow x^{2}+y^{2}-2x+2y=4 7

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