Prove that the circle $x^{2} + y^{2} + 2 a x + c^{2} = 0$ and $x^{2} + y^{2} + 2 b y + c^{2} = 0$ touches each other, if $\frac{1}{a^{2}} + \frac{1}{b^{2}} = \frac{1}{c^{2}}$ .

Open in App

Solution

$(x + a)^{2} + y^{2} = {(\sqrt{a^{2} - c^{2}})}^{2} x^{2} + (y + b)^{2} = {(\sqrt{b^{2} - c^{2}})}^{2}$
If these circles touch each other
distance between centres $=$ Sum of radii
$\sqrt{a^{2} + b^{2}} = \sqrt{a^{2} - c^{2}} + \sqrt{b^{2} - c^{2}} a^{2} + b^{2} = a^{2} - c^{2} + b^{2} - c^{2} + 2 \sqrt{a^{2} - c^{2}} \sqrt{b^{2} - c^{2}} 2 c^{2} = 2 \sqrt{a^{2} - c^{2}} \sqrt{b^{2} - c^{2}} c^{4} = (a^{2} - c^{2}) (b^{2} - c^{2}) c^{4} = a^{2} b^{2} - (b^{2} + a^{2}) c^{2} + c^{4} (a^{2} + b^{2}) c^{2} = a^{2} b^{2}$
Dividing both sides by $a^{2} b^{2} c^{2}$
$\frac{1}{a^{2}} + \frac{1}{b^{2}} = \frac{1}{c^{2}}$

Suggest Corrections

Similar questions

x^{2} + y^{2} + 2 a x + c = 0

x^{2} + y^{2} + 2 b y + c = 0

\frac{1}{c} = \frac{1}{a^{2}} + \frac{1}{b^{2}}

c_{1}, c_{2}

r_{1}, r_{2}

c_{1} + c_{2} = r_{1} \pm r_{2}

x^{2} + y^{2} - 2 a x + c^{2} = 0

x^{2} + y^{2} - 2 b y + c^{2} = 0

x^{2} + y^{2} + c^{2} = 2 a x

x^{2} + y^{2} + c^{2} - 2 b y = 0

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

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

The two circles $x^{2} + y^{2} + 2 a x + c = 0 a n d x^{2} + y^{2} + 2 b y + c = 0$ touch if $\frac{1}{a^{2}} + \frac{1}{b^{2}} =$

Join BYJU'S Learning Program