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

Question

The lines 2x − 3y = 5 and 3x − 4y = 7 are diameters of a circle of area 154 sq unit. Then, the equation of the circle is (A)  x2 + y2 + 2x - 2y = 51  (B)  x2 + y2 - 2x - 2y = 49  (C)  x2 + y2 + 2x + 2y = 47  (D)  x2 + y2 - 2x + 2y = 47

Tardigrade Admin · Accepted Answer

Key Idea The intersection of two diameters lines are the centre of circle.
The intersection of the lines

2 x - 3 y = 5

and

3 x - 4 y = 7

is

(1, - 1)

which is the coordinates of the centre.
Also, area of circle

= 154

= π r^{2} = 154

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

\Rightarrow r = 7

∴

Required equation of circle is

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

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

Q. The lines $2 x - 3 y = 5$ and $3 x - 4 y = 7$ are diameters of a circle of area $154 s q$ unit. Then, the equation of the circle is

$x^{2} + y^{2} + 2 x - 2 y = 51$

$x^{2} + y^{2} - 2 x - 2 y = 49$

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

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

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

x2+y2+2x−2y=51

x2+y2−2x−2y=49

x2+y2+2x+2y=47

x2+y2−2x+2y=47

Q. The lines $2 x - 3 y = 5$ and $3 x - 4 y = 7$ are diameters of a circle of area $154 s q$ unit. Then, the equation of the circle is

$x^{2} + y^{2} + 2 x - 2 y = 51$

$x^{2} + y^{2} - 2 x - 2 y = 49$

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

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