Correct Answer - Option 1 : x
2 + y
2 + 6x - 4y - 36 = 0
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
\(\rm (x-h)^2 + (y-k)^2 =R^2\)
⇒ \(\rm (x-(-3))^2+(y-2)^2 =7^2\)
⇒ \(\rm (x+3)^2+(y-2)^2 =49\)
⇒ \(\rm x^2+9+6x+y^2 +4 -4y-49=0\)
⇒ \(\boldsymbol{\rm x^2+6x+y^2-4y-36 =0}\)
