Question

the equation of circle of area 154 square unit two of whose diameter are 2 x -3 y+ 12 =0 and X + 4 Y - 5 =0 is

Answer 1

equation of diameters are 2x - 3y + 12 = 0 and x + 4y -5 = 0 .
Let ( h, k) is the centre of circle .
we know ,
all diameters pass through centre of circle .
hence, given equation of diameters too pass through circle.
now, put ( h, k) in both given equation .
2h - 3k + 12 = 0--------(1)
h + 4k - 5 = 0 ----------(2)

solve equations (1) and (2)
11k -22 = 0 => k = 2 and h = -3
hence, centre of circle is (2, -3)

now, area of circle is 154
Let radius of circle is r .
area of circle = πr²
154 = 22/7 r²
r = 7 unit

now, equation of circle when centre is (h,k) and radius is r :
(x - 2)² + (y + 3)² = 7²
x² + y² - 4x + 6y + 13 -49 = 0
x² + y² - 4x + 6y -36 = 0

Answer 2

Answer:

upper one is correct guys