Question

center and radius of 6x+x^2=97+10y-y^2

Answer 1

Answer:

• center of the circle = (-3, 5)
• radius of the circle = √131

Step-by-step explanation:

To find the center and radius of 6x + x² = 97 + 10y - y², we convert the equation of circle to the equation with center (a, b) and radius r:

$\boxed{(x-a)^2+(y-b)^2=r^2}$

$6x + x^2 = 97 + 10y - y^2$

$x^2+y^2+6x-10y=97$

$x^2+y^2+2(3x)-2(5y)=97$

$[(x+3)^2-3^2]+[(y-5)^2-5^2]=97$

$(x+3)^2+(y-5)^2=97+3^2+5^2$

$(x+3)^2+(y-5)^2=\sqrt{131} ^2$

By comparing $(x+3)^2+(y-5)^2=\sqrt{131} ^2$ with $(x-a)^2+(y-b)^2=r^2$, we can find that:

• $a=-3$
• $b=5$
• $r=\sqrt{131}$

Therefore:

• center of the circle = (-3, 5)
• radius of the circle = √131