Radius of circle touching y-axis at point $P (0, 2)$ and circle $x^{2} + y^{2} = 16$ internally-

\frac{5}{2}

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

\frac{3}{2}

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

\frac{5}{4}

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

2

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

Open in App

Solution

The correct option is C $2$
$\begin{matrix} D i a m e t e r o f c i r c l e t o u c h y - a x i s a t P (0, 2) a n d c i r c l e x^{2} + y^{2} = 16 W h e n c i r c l e t o u c h e s i n t e r n a l l y | c_{1} c_{2} | = | r_{1} - r_{2} | f r o m t h e f i g u r e w e c a n s a y t h a t ∴ D i a m e t e r o f t h e c i r c l e i s 2 H e n c e, t h e o p t i o n D i s t h e c o r r e c t a n s w e r . \end{matrix}$

Suggest Corrections

Similar questions

x^{2} + y^{2} = 16

x^{2} + y^{2} = 5

x^{2} + y^{2} - 8 x + 6 y + 20 = 0

x^{2} + y^{2} - 8 x - 4 y + 16 = 0

(4, 0)

(x - x_{1})^{2} + (y - r)^{2} = r^{2}

(x_{1} 0)

3

x^{2} + y^{2} - 2 x - 4 y - 20 = 0

(5, 5)

Equation of the circle whose radius is 3 and which touches the circle $x^{2} + y^{2} - 4 x - 6 y - 12 = 0$ internally at the point $(- 1, - 1)$ , is

Join BYJU'S Learning Program