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NIMCET 2016 Official Paper

Option 1 : x^{2} + y^{2} + 6x - 4y - 36 = 0

NIMCET 2020 Official Paper

1447

120 Questions
480 Marks
120 Mins

__Concept:__

The standard form of the equation of a circle is:

\(\rm (x-h)^2 + (y-k)^2 =R^2\)

where (h,k) are the coordinates and the R is the radius of center of the circle

Area of the circle = π R2

__Note:__ The intersection of the Equation of diameters is center of the circle

__Calculation:__

Given area of circle = 154 sq.units

⇒ π R2 = 154

⇒ R2 = \(\rm 154* \frac{7}{22}\)

⇒ R = 7

Equation of the diameters

__\(\rm 2x-3y+12=0\)__ ...(i)

\(\rm x+4y-5=0\) ...(ii)

Intersection of the diameters \(\boldsymbol{\rm (i)-2*(ii)}\)

⇒ \(\rm -11y+22=0\)

⇒ \(\boldsymbol{\rm y=2}\)

Putting back in equation (i),

⇒ \(\rm 2x-3(2)+12=0\)

⇒ \(\rm 2x=6 ⇒ \boldsymbol{\rm x=-3}\)

The center will be (-3, 2)

By the standard equation of circle

\(\rm (x-h)^2 + (y-k)^2 =R^2\)

⇒ \(\rm (x-(-3))^2+(y-2)^2 =7^2\)

⇒ \(\rm (x+3)^2+(y-2)^2 =49\)

⇒ \(\rm x^2+9+6x+y^2 +4 -4y-49=0\)

⇒ \(\boldsymbol{\rm x^2+6x+y^2-4y-36 =0}\)