Question

find the equation of a circle whose diameter are 2x-3y+12=0and x+4y-5=0and are 154square unit​

Answer 1

2x-3y = -12

x+4y = 5

2x-3y = -12

2x+8y = 10

-11y = -22

y = 2 ,

putting this in any equation we get , x = (-3)

so, centre of circle = (-3,2)

Now

πr² = 154

r² = 154*7/22 = 7*7

r = 7

so, equation of circle will be = (x+3)²+(y-2)² = (7)² (Ans.)

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