The diameters of a circle are along 2x+y-7-6 and x+3y-11-0. Then the equation of this circle which also passes through (5,7) is

Answers (1)

Let's assume that circle equation is (x-h)^2+(y-k)^2=r^2
Where, (h,k)
ightarrow centre
By solving the diameter we get center as diameters passes through centre 
Subtracting the equation 2x+6y-22=0 from 2x+y-7=0,we get
-5y+15=0Rightarrow y=3
Putting the value of y in one of the equation,we get
2x+3-7=0
2x-4=0
x=2
	herefore Centre(2,3)
Point on the circle is (5,7)
Radius between centre and point
sqrt (x_2-x_1)^2+(y_2-y_1)^2
sqrt(5-2)^2+(7-3)^2
=sqrt25=5
By substituting Centre (2,3) radius 5 in circle equation
(x-2)^2+(y-3)^2=5^2
x^2+y^2-4x-6y+13=25
x^2+y^2-4x-6y-12=0

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