The circles $x^{2} + y^{2} - 4 x + 6 y + 8 = 0$ and $x^{2} + y^{2} - 10 x - 6 y + 14 = 0$

touch externally

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

touch internally

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

intersect

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

do not meet

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

Open in App

Solution

The correct option is A touch externally
[ From point (1) ]
$S_{1} : x^{2} + y^{2} - 4 x + 6 y + 8 = 0$

$C_{1} \equiv (2, - 3)$ and $r a d i u s = r_{1} = \sqrt{5}$
$S_{2} : x^{2} + y^{2} - 10 x - 6 y + 14 = 0$
$C_{2} \equiv (5, 3)$ and $r_{2} = \sqrt{20} = 2 \sqrt{5}$
$∴ C_{1} C_{2} = \sqrt{5} = r_{1} + r_{2} = 3 \sqrt{5}$
$C_{1} C_{2} = r_{1} + r_{2}$
$∴ S_{1}$ and $S_{2}$ touching externally.

Suggest Corrections

Similar questions

x^{2} + y^{2} - 4 x - 6 y - 14 = 0

x^{2} + y^{2} + 2 x + 4 y - 5 = 0

x^{2} + y^{2} - 10 x - 16 y + 7 = 0

x^{2} + y^{2} - 4 x - 6 y - 14 = 0

x^{2} + y^{2} + 2 x + 4 y - 5 = 0

x^{2} + y^{2} - 10 x - 16 y + 7 = 0

x^{2} + y^{2} - 6 x - 6 y + 14 = 0

x^{2} + y^{2} - 4 x - 6 y - 14 = 0

x^{2} + y^{2} + 2 x + 4 y - 5 = 0

x^{2} + y^{2} - 10 x - 16 y + 7 = 0

x^{2} + y^{2} - 2 x - 3 = 0

x^{2} + y^{2} - 4 x - 6 y - 8 = 0

Join BYJU'S Learning Program