1
TS EAMCET 2020 (Online) 14th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

In the List-I each item contains equations of two circles, List-II contains the number of common tangents for each pair of circles given in List-I. Match the items of List-I with those of the items of List-II

$$
\text { List-I }
$$
$$
\text { List-II }
$$
A. $$
\begin{aligned}
& x^2+y^2+2 x+8 y-23=0 \\
& x^2+y^2-4 x-10 y+19=0
\end{aligned}
$$
I. 0
B. $$
\begin{aligned}
& x^2+y^2=1 \\
& x^2+y^2-2 x-6 y+6=0
\end{aligned}
$$
II. 1
C. $$
\begin{aligned}
& x^2+y^2-8 x+2 y=0 \\
& x^2+y^2-2 x-16 y+25=0
\end{aligned}
$$
III. 2
D. $$
\begin{aligned}
& x^2+y^2=4 \\
& x^2+y^2-2 x=0
\end{aligned}
$$
IV. 3
V. 4

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

A
A B C D
V I III II
B
A B C D
IV I III II
C
A B C D
IV V III II
D
A B C D
III IV I IV
2
TS EAMCET 2020 (Online) 14th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

$\left(0, \frac{3}{4}\right)$ is the radical centre of the circles $S_1: x^2+y^2-2 x+6 y=0, S_2: x^2+y^2+2 g x-2 y+6=0$ and $S_3: x^2+y^2-12 x+2 f y+3=0$. If $S_2$ and $S_3$ intersect orthogonally, then $(g, f)=$

A

$\left(\frac{-11}{12}, 1\right)$

B

$\left(1, \frac{-21}{2}\right)$

C

$\left(0, \frac{-9}{2}\right)$

D

$\left(-1, \frac{-7}{12}\right)$

3
TS EAMCET 2020 (Online) 14th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

For the circles $(x-a)^2+y^2=a^2$ and $x^2+(y-a)^2=a^2$, where $a>0$, which one of the following is not true?

A

mid-point of the common chord is $\left(\frac{a}{2}, \frac{a}{2}\right)$

B

length of the common chord is $(\sqrt{2} a)$

C

the circles intersect at $(0,0)$ and $(a, a)$

D

common chord is at a distance of $(\sqrt{2} a)$ units from the centres of given circles

4
TS EAMCET 2020 (Online) 14th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

If $S(a, b)$ is a fixed point and $P(\alpha, \beta)$ is such a variable point that $4\left[(x-a)^2+(y-b)^2\right]=(\alpha x+\beta y+7)^2$ represents a parabola, then the locus of $P(\alpha, \beta)$ is

A

$\beta^2=4 \alpha$

B

$\alpha^2+\beta^2=4$

C

$\frac{\alpha^2}{4}+\frac{\beta^2}{2}=1$

D

$(\alpha+\beta)^2=4$

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