Question

Determine the center and radius of the following circle equation:
x2 + y2 + 16x + 14y + 97 = 0

Answer 1

Answer: (-8,-7) is the center of circle with radius 4 units

Explanation:

Given equation :
`x^2+y^2+16x+14y+97=0`

We have to find the center and radius of the circle

As we know the equation of circle is of the form

`(x-h)^2+(y-k)^2=r^2` where (h,k) is center and r is radius

So we have
`x^2+y^2+16x+14y+97=0`

Using completing the square method

`x^2+16x+y^2+14y+97=0\\\\\Rightarrow x^2+16x+64-64+y^2+14y+49-49+97=0\\\\\Righatrrow (x^2+16x+64)+(y^2+14y+49)-49-64+97=0\\\\\Righatrrow (x+8)^2+(y+7)^2-16=0 \text { }[{\because}(a+b)^2=a^2+b^2+2ab ]\\\\\Righatrrow (x+8)^2+(y+7)^2=16\\\\\Righatrrow (x+8)^2+(y+7)^2=4^2`

Comparing to the equation of circle we get

h = -8 , k = -7 and r = 4

Hence, (-8,-7) is the center of circle with radius 4 units