How do you write the standard equation for the circle passing through pts (8,-2), (6,2) and (3,-7)?

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Answer 1 · 2017-12-31T13:05:38

The equation of circle is $(x-3)^2+(y+2)^2=25$

Let the points are $P(8,-2) , Q( 6,2) ,R(3,-7)$ and Centre of

the circle is $(a,b)$ and radius is $r$

The distance from center to the given 3 points are equal.

$(8-a)^2+(-2-b)^2=(6-a)^2+(2-b)^2=(3-a)^2+(-7-b)^2$

$(8-a)^2+(-2-b)^2=(6-a)^2+(2-b)^2$

$:. 64-16a+cancela^2+4+4b+cancelb^2= 36-12a+cancela^2+4-4b+cancelb^2$ or

$-4a+8b=-28 or 2b-a= -7(1)$ Similarly,

$(8-a)^2+(-2-b)^2=(3-a)^2+(-7-b)^2$

$:. 64-16a+cancela^2+4+4b+cancelb^2= 9-6a+cancela^2+49+14b+cancelb^2$ or

$-10a-10b= =-10 or a+b=1(2)$ Adding equation(1) and (2) we

get, $3b=-6 or b =-2:. a =1-b=1-(-2)=3 :.$

Centre is $(3,-2)$ Radius is $r=|PC|=sqrt ((8-a)^2+(-2-b)^2)$

$:. r= sqrt ((8-3)^2+(-2+2)^2)=5$

The equation of circle is $(x-a)^2+(y-b)^2=r^2$ or

$(x-3)^2+(y+2)^2=25$ [Ans]