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

Accumulation of which of the following molecules in the muscles occurs as a result of vigorous exercise?

The molecular formula of compound $A$ is $C_2Cl_3OH$. It reduces Fehling's solution and on oxidation gives a monocarboxylic acid $B$. Compound $A$ is obtained by the reaction of chlorine with ethyl alcohol. What is $A$?

An eight-digit number divisible by $9$ is to be formed using digits from $0$ to $9$ without repeating the digits. The number of ways in which this can be done is: (in $(7!)$)

The coefficient of $x^{24}$ in the expansion of $(1+x^2)^{12}(1+x^{12})(1+x^{24})$ is

Which of the following is the correct order of increasing solubility for the given compounds?

Vedclass Test Series

Exam Paper Generator

Online Exam Module