Answer: (-8,-7) is the center of circle with radius 4 units
Step-by-step explanation:
Given equation : [tex]x^2+y^2+16x+14y+97=0[/tex]
We have to find the center and radius of the circle
As we know the equation of circle is of the form
[tex](x-h)^2+(y-k)^2=r^2[/tex] where (h,k) is center and r is radius
So we have [tex]x^2+y^2+16x+14y+97=0[/tex]
Using completing the square method
[tex]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[/tex]
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
