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

Name: 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
Uploaded: Mon, 2021-05-24 14:33
Description: 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

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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
$(x-2)^2+(y-3)^2=5^2$
$x^2+y^2-4x-6y+13=25$
$x^2+y^2-4x-6y-12=0$