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 first circle $x^2 + y^2 - 2x - 4y = 0$,the center $C_1 = (1, 2)$ and radius $R_1 = \sqrt{1^2 + 2^2 - 0} = \sqrt{5}$.For the second circle

Vedclass · Accepted Answer

Touch each other internally

Vedclass · Answer

(A) For the first circle $x^2 + y^2 - 2x - 4y = 0$,the center $C_1 = (1, 2)$ and radius $R_1 = \sqrt{1^2 + 2^2 - 0} = \sqrt{5}$.For the second circle $x^2 + y^2 - 8y - 4 = 0$,the center $C_2 = (0, 4)$ and radius $R_2 = \sqrt{0^2 + 4^2 - (-4)} = \sqrt{20} = 2\sqrt{5}$.The distance between the centers $C_1$ and $C_2$ is $C_1C_2 = \sqrt{(1-0)^2 + (2-4)^2} = \sqrt{1^2 + (-2)^2} = \sqrt{1+4} = \sqrt{5}$.We observe that $|R_2 - R_1| = |2\sqrt{5} - \sqrt{5}| = \sqrt{5}$.Since $C_1C_2 = |R_2 - R_1|$,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

The slope of one of the direct common tangents drawn to the circles $x^2+y^2-2x+4y+1=0$ and $x^2+y^2-4x-2y+4=0$ is

If the circles $x^2+y^2-6x-8y-12=0$ and $x^2+y^2-4x+6y+k=0$ are orthogonal to each other,then the value of $k$ is:

$(a, 0)$ and $(b, 0)$ are the centres of two circles belonging to a coaxial system of which the $y$-axis is the radical axis. If the radius of one of the circles is $r$,then the radius of the other circle is

The radical centre of the circles $x^2+y^2-4x-6y+5=0$,$x^2+y^2-2x-4y-1=0$ and $x^2+y^2-6x-2y=0$ is equal to

The equation of the circle described on the chord $3x + y + 5 = 0$ of the circle $x^2 + y^2 = 16$ as diameter is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module