Question

If the lines 2x-3y=5 and 3x-4y=7 are the diameters of a circle of area 154 square units, then obtain the equation of the circle.
A. `x^(2)+y^(2)+2x-2y=62`
B. `x^(2)+y^(2)+2x-2y=47`
C. `x^(2)+y^(2)-2x+2y=47`
D. `x^(2)+y^(2)-2x+2y=62`

Answer 1

Correct Answer - C
Since, 2x-3y = 5 and 3x-4y = 7 are diameters of a circle.
Their point of intersection is centre (1, -1)
Are given, `pir^(2) = 154`
`rArr r^(2) = 154 xx (7)/(22) rArr r = 7 `
` therefore` Required of circle is
`(x-1)^(2) + (y+1)^(2) = 7^(2)`
`rArr x^(2) + y^(2)-2x +2y = 47`