1
TS EAMCET 2020 (Online) 10th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

The circle $S=0$ cuts the circles

$C_1=x^2+y^2-8 x-2 y+16=0$ and $C_2=x^2+y^2-4 x-4 y-1=0$ orthogonally. If the common chord of $S=0$ and $C_1=0$ is $2 x+13 y-15=0$, then the centre of $S=0$ is

A

$\left(\frac{-11}{3}, \frac{7}{6}\right)$

B

$\left(\frac{11}{3}, \frac{-7}{6}\right)$

C

$\left(\frac{2}{13}, \frac{11}{15}\right)$

D

$\left(\frac{11}{15}, \frac{-2}{13}\right)$

2
TS EAMCET 2020 (Online) 10th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

The equation of the circle passing through the points of intersection of the two orthogonal circles $S_1=x^2+y^2+k x-4 y-1=0$, $S_2=3 x^2+3 y^2-14 x+23 y-15=0$ and passing through the point $(-1,-1)$ is

A

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

B

$3 x^2+3 y^2+18 x-12 y=0$

C

$5 x^2+5 y^2-22 x+15 y-17=0$

D

$x^2+y^2-5 x+14 y+7=0$

3
TS EAMCET 2020 (Online) 10th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

Consider the parabola $y^2+2 x+2 y-3=0$ and match the items of List-I with those of the List-II.

$$ \begin{array}{llll} \hline & \text { List-I } & & \text { List-II } \\ \hline \text { A. } & 2 x-5=0 & \text { I. } & \text { Vertex } \\ \hline \text { B. } & \left(\frac{3}{2},-1\right) & \text { II. } & \text { Focus } \\ \hline \text { C. } & y+1=0 & \text { III. } & \text { Equation of directrix } \\ \hline \text { D. } & (2,-1) & \text { IV. } & \text { Equation of the axis } \\ \hline & & \text { V. } & \text { Equation of the Latus rectum } \\ \hline \end{array} $$

$$ \text { The correct match is } $$

A
A B C D
III II IV I
B
A B C D
V I IV II
C
A B C D
III II IV I
D
A B C D
IV I III II
4
TS EAMCET 2020 (Online) 10th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

The normal at a point on the parabola $y^2=4 x$ passes through $(5,0)$. If there are two more normals to this parabola which pass through $(5,0)$, the centroid of the triangle formed by the feet of these three normals is

A

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

B

$(4,0)$

C

$(0,2)$

D

$(2,0)$

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