Find the radius of circle touching $x$ -axis and concentric with $x^{2} + y^{2} - 2 x - 2 y + 3 = 0$ .

2

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

1

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

\sqrt{2}

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

2 \sqrt{2}

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

Open in App

Solution

The correct option is B $1$
Center of given circle is $C (1, 1)$

We need to find radius of the circle which has center C and has X- Axis as tangent

Radius = perpendicular distance from center to tangent $(y = 0)$

$r = \frac{| 1 |}{\sqrt{1^{2}}} = 1$

Suggest Corrections

Similar questions

x^{2} + y^{2} - 2 x - 2 y + 1 = 0

x^{2} + y^{2} + 2 x - 2 y + 1 = 0

2

x + y = 2 + \sqrt{2}

x^{2} + y^{2} - 2 x - 2 y + 1 = 0

The equation of a circle of radius 1 touching the circles $x^{2} + y^{2} - 2 | x | = 0$ is

\frac{x^{2}}{7} + \frac{y^{2}}{3} = 1

x^{2} + y^{2} + 2 g x + 2 f y + c = 0

2 x^{2} + 2 y^{2} + 3 x + 8 y + 2 c = 0

x^{2} + y^{2} + 2 x + 2 y + 1 = 0

g = \frac{3}{4}

f = 2

Join BYJU'S Learning Program