The equation of a circle with diameters are 2x - 3y + 12 = 0 and x + 4

1.	The equation of a circle with diameters are 2x - 3y + 12 = 0 and x + 4y - 5 = 0 and area of 154 sq. units is1. x2 + y2 + 6x - 4y - 36 = 02. x2 + y2 + 6x + 4y - 36 = 03. x2 + y2 - 6x + 4y + 25 = 04. None of these
Answer» Correct Answer - Option 1 : x² + y² + 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 \(\rm (x-h)^2 + (y-k)^2 =R^2\) ⇒ \(\rm (x-(-3))^2+(y-2)^2 =7^2\) ⇒ \(\rm (x+3)^2+(y-2)^2 =49\) ⇒ \(\rm x^2+9+6x+y^2 +4 -4y-49=0\) ⇒ \(\boldsymbol{\rm x^2+6x+y^2-4y-36 =0}\)

The equation of a circle with diameters are 2x - 3y + 12 = 0 and x + 4y - 5 = 0 and area of 154 sq. units is1. x2 + y2 + 6x - 4y - 36 = 02. x2 + y2 + 6x + 4y - 36 = 03. x2 + y2 - 6x + 4y + 25 = 04. None of these

Answer» Correct Answer - Option 1 : x² + y² + 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

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

⇒ \(\rm (x-(-3))^2+(y-2)^2 =7^2\)

⇒ \(\rm (x+3)^2+(y-2)^2 =49\)

⇒ \(\rm x^2+9+6x+y^2 +4 -4y-49=0\)

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