Two orthogonal circles are such that area of one is twice the area of other. If radius of smaller circle is $r$, then distance between their centers will be -
Two orthogonal circles are such that area of one is twice the area of other. If radius of smaller circle is $r$, then distance between their centers will be -
- A
$\sqrt 3 r$
- B
$2r$
- C
$\sqrt 5 r$
- D
$3r$
Similar Questions
The equation of the circle which touches the circle ${x^2} + {y^2} - 6x + 6y + 17 = 0$ externally and to which the lines ${x^2} - 3xy - 3x + 9y = 0$ are normals, is
The equation of the circle which touches the circle ${x^2} + {y^2} - 6x + 6y + 17 = 0$ externally and to which the lines ${x^2} - 3xy - 3x + 9y = 0$ are normals, is
If the circles ${x^2} + {y^2} = 4,{x^2} + {y^2} - 10x + \lambda = 0$ touch externally, then $\lambda $ is equal to
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A circle $S$ passes through the point $(0,1)$ and is orthogonal to the circles $(x-1)^2+y^2=16$ and $x^2+y^2=1$. Then
$(A)$ radius of $S$ is $8$
$(B)$ radius of $S$ is $7$
$(C)$ centre of $S$ is $(-7,1)$
$(D)$ centre of $S$ is $(-8,1)$
A circle $S$ passes through the point $(0,1)$ and is orthogonal to the circles $(x-1)^2+y^2=16$ and $x^2+y^2=1$. Then
$(A)$ radius of $S$ is $8$
$(B)$ radius of $S$ is $7$
$(C)$ centre of $S$ is $(-7,1)$
$(D)$ centre of $S$ is $(-8,1)$
- [IIT 2014]
Consider the circles ${x^2} + {(y - 1)^2} = $ $9,{(x - 1)^2} + {y^2} = 25$. They are such that
Consider the circles ${x^2} + {(y - 1)^2} = $ $9,{(x - 1)^2} + {y^2} = 25$. They are such that
The centre$(s)$ of the circle$(s)$ passing through the points $(0, 0) , (1, 0)$ and touching the circle $x^2 + y^2 = 9$ is/are :
The centre$(s)$ of the circle$(s)$ passing through the points $(0, 0) , (1, 0)$ and touching the circle $x^2 + y^2 = 9$ is/are :