Find the centre and radius of the circle $x^{2} + y^{2} - 8 x + 10 y - 12 = 0$ .

(4, - 5), \sqrt{53}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

(- 4, - 5), \sqrt{53}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(- 4, 5), \sqrt{53}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(4, 5), \sqrt{53}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $(4, - 5), \sqrt{53}$
Given circle equation
$x^{2} + y^{2} - 8 x + 10 y - 12 = 0 h e r e 2 g = - 8 ∴ g = - 4 a n d 2 f = 10 ∴ f = 5 ∴ c e n t r e (- g, - f) = (4, - 5) r a d i u s = \sqrt{g^{2} + f^{2} - c} = \sqrt{4^{2} + 5^{2} - (- 12)} = \sqrt{16 + 25 + 12} = \sqrt{53} u n i t$

Suggest Corrections

Similar questions

{(\frac{5}{3})}^{- 4} \times {(\frac{5}{3})}^{- 5} = {(\frac{5}{3})}^{3 x} .

\frac{4}{5}

\frac{3}{5}

(4, 5)

x^{2} + y^{2} + 4 x + 6 y - 12 = 0

x^{2} + y^{2} + 8 x + 10 y - 8 = 0

(\frac{5^{- 4}}{5^{- 6}}) \times 5^{3}

Join BYJU'S Learning Program