Question

If the circles $x^2+y^2+2ax+c=0and{x}^2+y^2+2by+c=0$ touch each other, then c is _________.where $a\&bϵR$.

• A. Always Negative
• B. Always positive
• C. Zero
• D. None of these

Answer 1

The correct option is B

Always positive

Given circles,

$x^2+y^2+2ax+c=0$ -------------(1)

$x^2+y^2+2by-c=0$ --------------(2)

Let $c_1\&c_2$ be tje centers and $r_1\&r_2$ radii of the circles (1) and (2) respectively.

$c_1(-a,0)$

$c_2(0,-b)$ distance between two centers $c_1.c_2=\sqrt{a^2+b^2}$

$r_1=\sqrt{g^2+f^2-c}=\sqrt{a^2-c}$

$r_2=\sqrt{g^2+f^2-c}=\sqrt{b^2-c}$

Case1: If circles touch each other, internally

Difference between the centers of the circles i.e., (-a,0) and (0,-b)= Difference of the radii of the circles

Case:2 If circles touch each other, externally

Difference between the centers of the circle i.e., (-a,0) and (0,-b)= sum of the radii of the circles

Overall we can saydistance between the two circles= sum or difference of radii of the circles

$c_1.c_2=|r_1\pm r_2|$

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

squaring on both sides

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

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

Again, squaring on both sides

$c^2=(a^2-c)(b^2-c)$

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

$(a^2+b^2)c=a^2{b}^2$

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

$a^2\&b^2$ will always be positive.

So, c is always positive.