The circles x^2 + y^2 - 2x - 4y = 0 and x^2 + | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) For the circle $x^2 + y^2 + 2gx + 2fy + c = 0$,the center is $(-g, -f)$ and the radius is $r = \sqrt{g^2 + f^2 - c}$.For the first circle $C_1: x^

Vedclass · Accepted Answer

Touch each other internally.

Vedclass · Answer

(A) For the circle $x^2 + y^2 + 2gx + 2fy + c = 0$,the center is $(-g, -f)$ and the radius is $r = \sqrt{g^2 + f^2 - c}$.For the first circle $C_1: x^2 + y^2 - 2x - 4y = 0$:Center $C_1 = (1, 2)$,Radius $r_1 = \sqrt{1^2 + 2^2 - 0} = \sqrt{5}$.For the second circle $C_2: x^2 + y^2 - 8y - 4 = 0$:Center $C_2 = (0, 4)$,Radius $r_2 = \sqrt{0^2 + 4^2 - (-4)} = \sqrt{16 + 4} = \sqrt{20} = 2\sqrt{5}$.The distance between the centers $d = \sqrt{(1 - 0)^2 + (2 - 4)^2} = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}$.We observe that $|r_2 - r_1| = |2\sqrt{5} - \sqrt{5}| = \sqrt{5}$.Since the distance between the centers $d = |r_2 - r_1|$,the circles touch each other internally.

The circles $x^2 + y^2 - 2x - 4y = 0$ and $x^2 + y^2 - 8y - 4 = 0$:

Similar Questions

If $\cos (A-B)=3/5$ and $\tan A \tan B=2$,then which one of the following is true?

$D$-glucose and $D$-fructose can be differentiated by

If an electron and a proton have the same de-Broglie wavelength,then the kinetic energy of the electron is

$A$ ball of mass $160\, g$ is thrown up at an angle of $60^\circ$ to the horizontal at a speed of $10\, m\,s^{-1}$. The angular momentum of the ball at the highest point of the trajectory with respect to the point from which the ball is thrown is nearly ........ $kg\, m^2/s$ $(g = 10\, m\,s^{-2})$.

The polio virus is a ..........

Vedclass Test Series

Exam Paper Generator

Online Exam Module