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

Which of the following is the most stable conformer of butane?

Two particles $A$ and $B$ move with constant velocities $\vec{v}_1$ and $\vec{v}_2$. At the initial moment,their position vectors are $\vec{r}_1$ and $\vec{r}_2$ respectively. The condition for their collision is

Resistance of a decimolar solution between two electrodes $0.02 \ m$ apart and $0.0004 \ m^2$ in area was found to be $50 \ \Omega$. Specific conductance $(k)$ is : ........... $S \ m^{-1}$

If $n(A) = 3$,$n(B) = 6$ and $A \subseteq B$,then the number of elements in $A \cup B$ is:

If $\int \sqrt{\frac{x}{a^3-x^3}} \, dx = g(x) + c$, then $g(x)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module