( x − a ) 2 + ( y − b ) 2 = r 2 i s t h e c i r c l e e q u a t i o n w i t h a r a d i u s r , c e n t e r e d a t ( a , b ) \left(x−a\right)^2+\left(y−b\right)^2=r^2\:\:\mathrm{is\:the\:circle\:equation\:with\:a\:radius\:r,\:centered\:at}\:\left(a,\:b\right) ( x − a ) 2 + ( y − b ) 2 = r 2 is the circle equation with a radius r , centered at ( a , b )
Rewriting 2 x 2 + 2 y 2 + 10 x = 2 y + 7 2x^2+2y^2+10x=2y+7 2 x 2 + 2 y 2 + 10 x = 2 y + 7 in form of standard circle equation
2 x 2 + 10 x + 2 y 2 − 2 y = 7 2x^2+10x+2y^2-2y=7 2 x 2 + 10 x + 2 y 2 − 2 y = 7
x 2 + 5 x + y 2 − y = 7 2 x^2+5x+y^2-y=\frac{7}{2} x 2 + 5 x + y 2 − y = 2 7
Writing the above equation in form of ( x − a ) 2 + ( y − b ) 2 = r 2 \left(x−a\right)^2+\left(y−b\right)^2=r^2 ( x − a ) 2 + ( y − b ) 2 = r 2
( x − ( − 5 2 ) ) 2 + ( y − 1 2 ) 2 = ( 10 ) 2 \left(x-\left(-\frac{5}{2}\right)\right)^2+\left(y-\frac{1}{2}\right)^2=\left(\sqrt{10}\right)^2 ( x − ( − 2 5 ) ) 2 + ( y − 2 1 ) 2 = ( 10 ) 2
( a , b ) = ( − 5 2 , 1 2 ) , r = 10 \left(a,\:b\right)=\left(-\frac{5}{2},\:\frac{1}{2}\right),\:r=\sqrt{10} ( a , b ) = ( − 2 5 , 2 1 ) , r = 10
C i r c l e w i t h c e n t e r a t ( − 5 2 , 1 2 ) a n d r a d i u s r = 10 \mathrm{Circle\:with\:center\:at}\:\left(-\frac{5}{2},\:\frac{1}{2}\right)\:\mathrm{and\:radius}\:r=\sqrt{10} Circle with center at ( − 2 5 , 2 1 ) and radius r = 10
Sketch its graph is below
Comments