Question

Two circles $x^2+y^2-2x-3=0$ and $x^2+y^2-4x-6y-8=0$ are such that:

• A. they touch each other
• B. they intersect each other
• C. one lies inside the other
• D. each lies outside the other

Answer 1

The correct option is B they intersect each other

Given circles are $x^2+y^2-2x-3=0$ -------(1)

and $x^2+y^2-4x-6y-8=0$ -------(2)

(1)-(2) is $⟹2x+6y+5=0$

$⟹y=\frac{-(2x+5)}{6}$ ------(3)

substituting (3) in (1) we get

$x^2+(\frac{-(2x+5)}{6}{)}^2-2x-3=0$

$⟹x^2+\frac{4x^2+20x+25}{6}-2x-3=0$

$⟹6x^2+4x^2+20x+25-12x-18=0$

$⟹10x^2+8x+7=0$

$⟹x=\frac{-8\pm \sqrt{-216}}{20}$

Therefore, the two circles intersect at two points.