The locus of centre of a circle passing through $(a, b)$ and cuts orthogonally to circle ${x^2} + {y^2} = {p^2}$, is
The locus of centre of a circle passing through $(a, b)$ and cuts orthogonally to circle ${x^2} + {y^2} = {p^2}$, is
- [AIEEE 2005]
- [IIT 1988]
- A
$2ax + 2by - ({a^2} + {b^2} + {p^2}) = 0$
- B
$2ax + 2by - ({a^2} - {b^2} + {p^2}) = 0$
- C
${x^2} + {y^2} - 3ax - 4by + ({a^2} + {b^2} - {p^2}) = 0$
- D
${x^2} + {y^2} - 2ax - 3by + ({a^2} - {b^2} - {p^2}) = 0$
Similar Questions
The condition that the circle ${(x - 3)^2} + {(y - 4)^2} = {r^2}$ lies entirely within the circle ${x^2} + {y^2} = {R^2},$ is
The condition that the circle ${(x - 3)^2} + {(y - 4)^2} = {r^2}$ lies entirely within the circle ${x^2} + {y^2} = {R^2},$ is
If the circles $x^{2}+y^{2}+6 x+8 y+16=0$ and $x^{2}+y^{2}+2(3-\sqrt{3}) x+x+2(4-\sqrt{6}) y$ $= k +6 \sqrt{3}+8 \sqrt{6}, k >0$, touch internally at the point $P(\alpha, \beta)$, then $(\alpha+\sqrt{3})^{2}+(\beta+\sqrt{6})^{2}$ is equal to $\dots\dots$
If the circles $x^{2}+y^{2}+6 x+8 y+16=0$ and $x^{2}+y^{2}+2(3-\sqrt{3}) x+x+2(4-\sqrt{6}) y$ $= k +6 \sqrt{3}+8 \sqrt{6}, k >0$, touch internally at the point $P(\alpha, \beta)$, then $(\alpha+\sqrt{3})^{2}+(\beta+\sqrt{6})^{2}$ is equal to $\dots\dots$
- [JEE MAIN 2022]
The distance from the centre of the circle $x^2 + y^2 = 2x$ to the straight line passing through the points of intersection of the two circles $x^2 + y^2 + 5x -8y + 1 =0$ and $x^2 + y^2-3x + 7y -25 = 0$ is-
The distance from the centre of the circle $x^2 + y^2 = 2x$ to the straight line passing through the points of intersection of the two circles $x^2 + y^2 + 5x -8y + 1 =0$ and $x^2 + y^2-3x + 7y -25 = 0$ is-
The number of integral values of $\lambda $ for which $x^2 + y^2 + \lambda x + (1 - \lambda )y + 5 = 0$ is the equation of a circle whose radius cannot exceed $5$ , is
The number of integral values of $\lambda $ for which $x^2 + y^2 + \lambda x + (1 - \lambda )y + 5 = 0$ is the equation of a circle whose radius cannot exceed $5$ , is
The equation of a circle passing through points of intersection of the circles ${x^2} + {y^2} + 13x - 3y = 0$ and $2{x^2} + 2{y^2} + 4x - 7y - 25 = 0$ and point $(1, 1)$ is
The equation of a circle passing through points of intersection of the circles ${x^2} + {y^2} + 13x - 3y = 0$ and $2{x^2} + 2{y^2} + 4x - 7y - 25 = 0$ and point $(1, 1)$ is
- [IIT 1983]