Question

Find the centre and radius of the circle x2+y2-8x+10y-12=0

Answer 1

The above equation can be written as:

Therefore, the centre of the circle=(4,-5)

rαdíuѕ

= $\sqrt{53}$

Answer 2

The equation of the given circle is x2 + y2 – 8x + 10y – 12 = 0.

x2 + y2 – 8x + 10y – 12 = 0

⇒ (x2 – 8x) + (y2 + 10y) = 12

⇒ {x2 – 2(x)(4) + 42} + {y2 + 2(y)(5) + 52}– 16 – 25 = 12

⇒ (x – 4)2 + (y + 5)2 = 53

⟹(−4)2+{−(−5)}2=(√53)2

which is of the form (x – h)2 + (y – k)2 = r2, where h = 4, k = – 5 and r = √53

Thus, the centre of the given circle is (4, –5), while its radius is √53.