Test Index

Circles Part 1

Section: Mathematics

© examsnet.com

Question : 100 of 100

Marks: +1, -0

The center of the circle passing through

(0,0)

(1,0)

and touching the circle

x^2+y^2=9

$\left(\frac{1}{2}, \frac{1}{2}\right)$
$\left(\frac{1}{2},-\sqrt{2}\right)$
$\left(\frac{3}{2}, \frac{1}{2}\right)$
$\left(\frac{1}{2}, \frac{3}{2}\right)$

Solution:

Let the required circle be

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

Since it passes through

(0,0)

and

(1,0)

\Rightarrow c=0

and

g=-\frac{1}{2}

Points

(0,0)

and

(1, 0)

lie inside the circle

x^2+y^2=9

,

so two circles touch internally

\Rightarrow c_1 c_2=r_1-r_2

\therefore \sqrt{g^2+f^2}=3-\sqrt{g^2+f^2}

\Rightarrow \sqrt{g^2+f^2}=\frac{3}{2}

\Rightarrow f^2=\frac{9}{4}-\frac{1}{4}=2

\therefore f=\pm \sqrt{2}

Hence, the centers of required circle are

\left(\frac{1}{2} \cdot \sqrt{2}\right)

or

\left(\frac{1}{2},-\sqrt{2}\right)

© examsnet.com

Go to Question:

Prev Question