Question

What would be the maximum Y value on the circle whose equation is shown below. x^2+10x+y^2-4y=71

Answer 1

Answer:

12.

Step-by-step explanation:

The given equation is

$x^2+10x+y^2-4y=71$

It can be written as

$(x^2+10x)+(y^2-4y)=71$

$(x^2+10x+5^2)+(y^2-4y+2^2)=71+5^2+2^2$

$(x+5)^2+(y-2)^2=71+25+4$

$(x+5)^2+(y-2)^2=10^2$     ...(i)

The standard form of a circle is

$(x-h)^2+(y-k)^2=r^2$     ...(ii)

where, (h,k) is center and r is radius.

From (i) and (ii), we get

$h=-5,k=2,r=10$

The center of circle is (-5,2) and radius is 10. So, the maximum value of y is

$Max(y)=k+r=2+10=12$

Therefore, the maximum value of y is 12.