TS EAMCET 2020 (Online) 10th September Evening Shift | Circle Question 174 | Mathematics

Suppose the circle $S: x^2+y^2+2 g x+2 f y+c=0$ cuts orthogonally the two circles $S^{\prime}: x^2+y^2-4 x-6 y+11=0$ and $S^{\prime \prime}: x^2+y^2-10 x-4 y+21=0$. If the centre of $S=0$ lies on the bisector of the angle between the positive coordinate axes, then $2 g+2 f+c=$

TS EAMCET 2020 (Online) 10th September Evening Shift

MCQ (Single Correct Answer)

-0

If the circle $S_1: x^2+y^2=16$ intersects another circle $S_2$ of radius 5 units such that the common chord is of maximum length and slope $\frac{3}{4}$, then the centre of the circle $S_2$ is

$\left(\frac{-9}{5}, \frac{12}{5}\right)$ or $\left(\frac{9}{5}, \frac{-12}{5}\right)$

$\left(\frac{7}{5}, \frac{-12}{5}\right)$ or $\left(\frac{-7}{5}, \frac{12}{5}\right)$

$\left(\frac{-9}{5}, \frac{-12}{5}\right)$ or $\left(\frac{9}{5}, \frac{12}{5}\right)$

$\left(\frac{12}{5}, \frac{9}{5}\right)$ or $\left(\frac{-12}{5}, \frac{-9}{5}\right)$

TS EAMCET 2020 (Online) 10th September Morning Shift

MCQ (Single Correct Answer)

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Two points from the set of concyclic points of the circle passing through $(1,1),(2,-1),(3,2)$ is

$\left(\frac{5}{2}+\sqrt{\frac{5}{2}}, \frac{1}{2}+\sqrt{\frac{5}{2}}\right),\left(\frac{5}{2}, \frac{1}{2}+\sqrt{\frac{5}{2}}\right)$

$\left(\frac{5}{2}+\sqrt{\frac{5}{2}}, \frac{1}{2}\right),\left(\frac{5+\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)$

$\left(\frac{5+\sqrt{5}}{2}, \frac{1+\sqrt{5}}{\sqrt{2}}\right),\left(\frac{5}{2}+\sqrt{\frac{5}{2}}+\frac{1+\sqrt{5}}{4}\right)$

$\left(\frac{5}{2}-\frac{\sqrt{5}}{2}, \frac{1}{2}-\frac{\sqrt{5}}{2}\right)\left(\frac{5}{2}-\frac{\sqrt{5}}{2}, \frac{1}{2}+\frac{\sqrt{5}}{2}\right)$

TS EAMCET 2020 (Online) 10th September Morning Shift

MCQ (Single Correct Answer)

-0

If the polar of a point $P$ with respect to a circle of radius $r$ which touches the coordinate axes and lies in the first quadrant is $x+2 y=4 r$, then the point $P$ is

$(r, 2 r)$

$(2 r, r)$

$(2 r, 3 r)$

$(-r, 4 r)$

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