If the circles $x^{2} + y^{2} = a$ and $x^{2} + y^{2} - 6 x - 8 y + 9 = 0$ , touch externally, then a=

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Solution

The correct option is A
1

$x^{2} + y^{2} = a \dots (1)$
and $x^{2} + y^{2} - 6 x - 8 y + 9 = 0 \dots (2)$

Let circles (1) and (2) touch each other at point P.

The centre of the circle $x^{2} + y^{2} = a, O i s (0, 0)$

The centre of the circle $x^{2} + y^{2} - 6 x - 8 y + 9 = 0$ , is (3, 4)

Also, radius of circle $(1) = \sqrt{a} = O P$

Radius of circle
$(2) = \sqrt{9 + 16 - 9} = 4 = C_{1} P$

From figure, we have :
$C_{O} = C_{1} P + O P$

$\Rightarrow 3^{2} + 4^{2} = 4 + \sqrt{a}$

$\Rightarrow 5 = 4 + \sqrt{a}$

$\Rightarrow a = 1$

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Similar questions

A

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

x^{2} + y^{2} - 6 x - 8 y + 9 = 0

a = 1

R

C_{1}, C_{2}

r_{1}, r_{2}

C_{1} C_{2} = r_{1} + r_{2}

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

x^{2} + y^{2} + 2 x + 4 y - 11 = 0

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

k =

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

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

k

x^{2} + y^{2} = 4

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

K =

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