Question

if the circles x2+y2+2a+c=0 and x2+y+2 by+c=0 touch each other then ??

Answer 1

u forgot mentioning 2ax .. i assumed it to be that so this is the ans if its 2ay the ans would be way different ....also it not mention they touch internally or extranally so if its a multiple choice ans question ... try doin r1 - r2 = c1c2

Answer 2

Concept: A circle is a shape made up of all points in a plane that are at the same distance from a central point. It's the curve drawn out by a point moving in a plane while keeping its distance from a particular point constant.

Given:

$x^{2} + y^{2} +2ax = c = 0 \\and\\x^{2} + y^{2} +2by +c = 0$ touch each other

To find: the correct condition out of the 4 options

Solution: Circles are

$C_{1} = x^{2} + y^{2} + 2ax + c = 0\\C_{2} = x^{2} + y^{2} + 2by + c = 0\\C_{1} (-a, 0 ) and \\r_1 = \sqrt{a^{2} - c} \\C_{2} (-b, 0) and \\r_2 = \sqrt{b^{2} - c}$

Since, circles touch each other

$(C_1C_2)^2 = (r_1 + r_2)^2\\a^{2} + b^{2} = \sqrt{a^{2} - c } \sqrt{b^{2}-c }$

Squaring both the sides,

$a^{2} + b^{2} = a^{2} - c + b^{2} -c +2\sqrt{a^{2}-c } \sqrt{b^{2}-c } \\- 2c = 2\sqrt{a^{2}-c } \sqrt{b^{2}-c }$

Cancelling 2 from both the sides, and then squaring

$c^2 = ({a^{2}-c }) ({b^{2}-c })\\c^2 = a^2b^2 -c(a^2 +b^2) + c^2\\c = \frac{a^{2}b^{2} }{a^2 +b^2} \\\frac{1}{c} = \frac{a^2 +b^2}{a^{2}b^{2}} \\ \\\frac{1}{c} = \frac{1}{a^{2} } +\frac{1}{b^{2} }$

Hence, option (a) is correct.

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