The coordinates of the radical centre of the three circles ${x^2} + {y^2} - 4x - 2y + 6 = 0,{x^2} + {y^2} - 2x - 4y -1 = 0,$${x^2} + {y^2} - 12x + 2y + 30 = 0$ are
- A$(6, 30)$
- B$(0, 6)$
- C$(3, 0)$
- DNone of these
Similar Questions
A circle $C$ of radius $2$ lies in the second quadrant and touches both the coordinate axes. Let $r$ be the radius of a circle that has centre at the point $(2,5)$ and intersects the circle $C$ at exactly two points. If the set of all possible values of $r$ is the interval $(\alpha, \beta)$, then $3 \beta-2 \alpha$ is equal to :
- [JEE MAIN 2025]
A circle ${C_1}$ of radius $2$ touches both $x$ - axis and $y$ - axis. Another circle ${C_2}$ whose radius is greater than $2$ touches circle ${C_1}$ and both the axes. Then the radius of circle ${C_2}$ is
A circle ${C_1}$ of radius $2$ touches both $x$ - axis and $y$ - axis. Another circle ${C_2}$ whose radius is greater than $2$ touches circle ${C_1}$ and both the axes. Then the radius of circle ${C_2}$ is
If the circles ${x^2} + {y^2} = {a^2}$and ${x^2} + {y^2} - 2gx + {g^2} - {b^2} = 0$ touch each other externally, then
If the circles ${x^2} + {y^2} = {a^2}$and ${x^2} + {y^2} - 2gx + {g^2} - {b^2} = 0$ touch each other externally, then
The locus of the mid points of the chords of the circle $C_1:(x-4)^2+(y-5)^2=4$ which subtend an angle $\theta_i$ at the centre of the circle $C_1$, is a circle of radius $r_i$. If $\theta_1=\frac{\pi}{3}, \theta_3=\frac{2 \pi}{3}$ and $r_1^2=r_2^2+r_3^2$, then $\theta_2$ is equal to
The locus of the mid points of the chords of the circle $C_1:(x-4)^2+(y-5)^2=4$ which subtend an angle $\theta_i$ at the centre of the circle $C_1$, is a circle of radius $r_i$. If $\theta_1=\frac{\pi}{3}, \theta_3=\frac{2 \pi}{3}$ and $r_1^2=r_2^2+r_3^2$, then $\theta_2$ is equal to
- [JEE MAIN 2023]
The two circles ${x^2} + {y^2} - 2x - 3 = 0$ and ${x^2} + {y^2} - 4x - 6y - 8 = 0$ are such that
The two circles ${x^2} + {y^2} - 2x - 3 = 0$ and ${x^2} + {y^2} - 4x - 6y - 8 = 0$ are such that