find the equation of the circle which passes through the points (2,-2) and (3,4) and whose centre lies on the line x+y=2
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step1:- let center of circle =(a, b)
but center lies x+y=2
so, a+b=2
b=2-a
now center (a,2-a)
step 2:- because (2,-2) and (3,4) passing through circle
so,
(a-2)^2+(2-a+2)^2=(a-3)^2+(2-a-4)^2
{circle have constant radius i.e distance from circumference to center is always equal}
now solve above equation
a=0.7
2-a=1.3
now radius =root{(2-0.7)^2+(-2-1.3)^2}
but center lies x+y=2
so, a+b=2
b=2-a
now center (a,2-a)
step 2:- because (2,-2) and (3,4) passing through circle
so,
(a-2)^2+(2-a+2)^2=(a-3)^2+(2-a-4)^2
{circle have constant radius i.e distance from circumference to center is always equal}
now solve above equation
a=0.7
2-a=1.3
now radius =root{(2-0.7)^2+(-2-1.3)^2}
abhi178:
can you solve my equation
bro i am not getting a=0.7
after solving it gives
a=1.5 &b=0.5
(a-2)^2+(4-a)^2=(a-3)^2+(a+2)^2
please again you calculate
answer a=0.7 and 2-a =1.3 I again found
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