10-1.Circle and System of Circles
hard

બે વર્તૂળો ${x^2} + {y^2} = ax$ અને${x^2} + {y^2} = {c^2}$ એકબીજા ને સ્પર્શે છે,તો .

A

$|a|=c$

B

$a=2c$

C

$|a|=2c$

D

$2|a|=c$

(AIEEE-2011)

Solution

The centres and radii are

$\left(x-\frac{a}{2}\right)^{2}+y^{2}=\frac{a^{2}}{4}, \quad x^{2}+y^{2}=c^{2}$

Centre $\left(\frac{a}{2}, 0\right)$ and $(0,0)$ and radius $=\frac{a}{2}$ and $c$

$\sqrt{\left(\frac{a}{2}\right)^{2}+(0-0)}=|| \frac{a}{2}|\pm c|$ 

$ \Rightarrow\left|\frac{a}{2}\right|=|| \frac{a}{2}|\pm c|$

$\Rightarrow\left|\frac{a}{2}\right|=c-\left|\frac{a}{2}\right|, \quad \therefore|a|=c$

Standard 11
Mathematics

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