The radical centre of the circles $x^2 + y^2 + 4x + 6y = 19$,$x^2 + y^2 = 9$,and $x^2 + y^2 - 2x - 2y = 5$ is:

  • A
    $(1, 1)$
  • B
    $(-1, 1)$
  • C
    $(1, -1)$
  • D
    $(0, 1)$

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Similar Questions

The equation of the circle passing through $(0,0)$ and cutting orthogonally the circles $x^2+y^2+6x-15=0$ and $x^2+y^2-8y-10=0$ is

In List-$I$,a pair of circles is given in $A$,$B$,$C$ and in List-$II$,the angle between those pairs of circles is given. Match the items from List-$I$ to List-$II$.
List-$I$ List-$II$
$(A)$ $(x-2)^2+y^2=2$,$(x-2)^2+(y-1)^2=1$ $I.$ $90^{\circ}$
$(B)$ $x^2+y^2-6x-6y+9=0$,$x^2+y^2-4x+4y-9=0$ $II.$ $135^{\circ}$
$(C)$ $x^2+y^2+4x-14y+28=0$,$x^2+y^2+4x-5=0$ $III.$ $60^{\circ}$
$IV.$ $30^{\circ}$

The correct matching is

$A$ 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 $(-8,1)$

The equation of the circle touching the line $2x+3y+1=0$ at the point $(1,-1)$ and orthogonal to the circle which has the line segment having end points $(0,-1)$ and $(-2,3)$ as diameter,is

The centre of the circle passing through the points of intersection of the circles $(x+3)^2+(y+2)^2=25$ and $(x-2)^2+(y-3)^2=25$ and cutting the circle $(x+1)^2+(y-2)^2=16$ orthogonally is

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