A circle has a center that falls on the line y = 2x +7 and passes through (4 ,7 ) and (1 ,2 ). What is the equation of the circle?

1 Answer
Shwetank Mauria
May 17, 2016

Equation of the circle is 169x^2+169y^2+130x-2106y+3237=0

Explanation:

The circle passes through line y=2x+7 and it also passes through (4,7) and (1,2). As such it should be equidistant from these two points. Hence,

(x-4)^2+(y-7)^2=(x-1)^2+(y-2)^2

or x^2-8x+16+y^2-14y+49=x^2-2x+1+y^2-4y+4

or -8x+2x-14y+4y+16+49-1-4=0

or -6x-10y+60=0 but as y=2x+7

-6x-10(2x+7)+60=0 or -26x-10=0

or x=-10/26=-5/13 and y=2xx-5/13+7=-10/13+7=81/13

Hence center of circle is (-5/13,81/13) and radius of circle will its distance with (1,2) or

r^2=(1+5/13)^2+(2-81/13)^2=(18/13)^2+(-55/13)^2=(324+3025)/169=3349/169

And equation of circle is

(x+5/13)^2+(y-81/13)^2=3349/169 multiplying by 169

(13x+5)^2+(13y-81)^2=3349 or

169x^2+130x+25+169y^2-2106y+6561=3349

or 169x^2+169y^2+130x-2106y+3237=0

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