Answer:
Answer:
\sf (x+14)^2+(y+5)^2=149(x+14)2+(y+5)2=149
Explanation:
Standard equation of a circle: \sf (x-a)^2+(y-b)^2=r^2(x−a)2+(y−b)2=r2
(where (a, b) is the center and r is the radius of the circle)
Substitute the given center (-14, -5) into the equation:
\sf \implies (x-(-14))^2+(y-(-5))^2=r^2⟹(x−(−14))2+(y−(−5))2=r2
\sf \implies (x+14)^2+(y+5)^2=r^2⟹(x+14)2+(y+5)2=r2
Now substitute the point (-7, 5) into the equation to find r²:
\sf \implies ((-7)+14)^2+(5+5)^2=r^2⟹((−7)+14)2+(5+5)2=r2
\sf \implies (7)^2+(10)^2=r^2⟹(7)2+(10)2=r2
\sf \implies 149=r^2⟹149=r2
Final equation:
\sf (x+14)^2+(y+5)^2=149(x+14)2+(y+5)2=149