The lines 2x - 3y - 5 = 0 and 3x - 4y = 7 are diameters of a circle of area 154 sq units, then the equation of the circle is

Question

The lines 2x - 3y - 5 = 0 and 3x - 4y = 7 are diameters of a circle of area 154 sq units, then the equation of the circle is (A) x2 + y2 + 2x - 2y - 62 = 0 (B) x2 + y2 + 2x - 2y - 47 = 0 (C) x2 + y2 - 2x + 2y - 47 = 0 (D) x2 + y2 - 2x + 2y - 62 = 0

Tardigrade Admin · Accepted Answer

The centre of the required circle lies at the intersection of

2 x - 3 y - 5 = 0

and

3 x - 4 y - 7 = 0

.
Thus, the coordinates of the centre are

(1, - 1)

.
Let r be the radius of the circle.
Then

π r^{2} = 154

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

\Rightarrow r = 7

Hence, the equation of required circle is

(x - 1)^{2} + (y + 1)^{2} = 7^{2}

\Rightarrow x^{2} + y^{2} - 2 x + 2 y - 47 = 0

Q. The lines $2 x - 3 y - 5 = 0$ and $3 x - 4 y = 7$ are diameters of a circle of area 154 sq units, then the equation of the circle is

$x^{2} + y^{2} + 2 x - 2 y - 62 = 0$

$x^{2} + y^{2} + 2 x - 2 y - 47 = 0$

$x^{2} + y^{2} - 2 x + 2 y - 47 = 0$

$x^{2} + y^{2} - 2 x + 2 y - 62 = 0$

Q. The lines 2x−3y−5=0 and 3x−4y=7 are diameters of a circle of area 154 sq units, then the equation of the circle is

x2+y2+2x−2y−62=0

x2+y2+2x−2y−47=0

x2+y2−2x+2y−47=0

x2+y2−2x+2y−62=0

Q. The lines $2 x - 3 y - 5 = 0$ and $3 x - 4 y = 7$ are diameters of a circle of area 154 sq units, then the equation of the circle is

$x^{2} + y^{2} + 2 x - 2 y - 62 = 0$

$x^{2} + y^{2} + 2 x - 2 y - 47 = 0$

$x^{2} + y^{2} - 2 x + 2 y - 47 = 0$

$x^{2} + y^{2} - 2 x + 2 y - 62 = 0$