TS EAMCET 2020 (Online) 11th September Morning Shift | Circle Question 168 | Mathematics

If a circle of radius $r$ touches the positive coordinate axes and also the circle $x^2+y^2-12 x-10 y+52=0$ externally, then the distance between the centres of the two circles is

TS EAMCET 2020 (Online) 11th September Morning Shift

MCQ (Single Correct Answer)

-0

If the circles $x^2+y^2-2 x-2 y+k=0$ and $x^2+y^2+4 x+6 y+4=0$ touch each other externally, then the point of contact of the two circles is

$\left(\frac{-1}{5}, \frac{-3}{5}\right)$

$\left(\frac{-1}{3}, \frac{-1}{3}\right)$

$(-1,-3)$

$(-1,-1)$

TS EAMCET 2020 (Online) 11th September Morning Shift

MCQ (Single Correct Answer)

-0

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

$\left(\frac{-27}{2}, \frac{-25}{2}\right)$

$(0,0)$

$\left(\frac{16}{3}, \frac{-25}{4}\right)$

$\left(\frac{4}{7}, \frac{3}{7}\right)$

TS EAMCET 2020 (Online) 10th September Evening Shift

MCQ (Single Correct Answer)

-0

If the origin lies on a diameter of the circle $x^2+y^2-4 x-2 y-4=0$, then the equation of the circle passing through the end points of that diameter and the point $(1,2)$ is

$x^2+y^2-2 x-4 y=0$

$3 x^2+3 y^2-19 x+8 y-12=0$

$7 x^2+7 y^2-31 x-28 y+17=0$

$x^2+y^2=5$

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