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

The centre and radius of the circumcircle of the triangle formed by the lines $2 x+3 y=10, y=x$ and the $X$-axis are respectively.

A

$\left(\frac{-5}{2}, \frac{3}{2}\right), \frac{\sqrt{34}}{2}$

B

$\left(\frac{5}{2}, 2\right), \frac{\sqrt{41}}{2}$

C

$\left(\frac{5}{2}, \frac{-1}{2}\right), \sqrt{\frac{13}{2}}$

D

$\left(\frac{1}{2}, \frac{-5}{2}\right), \sqrt{\frac{13}{2}}$

2
TS EAMCET 2020 (Online) 14th September Morning Shift
MCQ (Single Correct Answer)
+1
-0

If the straight lines $3 x-4 y+4=0$ and $6 x-8 y-7=0$ are the tangents to the same circle, then the area of that circle (in square units) is

A

$\frac{9 \pi}{4}$

B

$\frac{9 \pi}{16}$

C

$\frac{25 \pi}{9}$

D

$\frac{121 \pi}{25}$

3
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
4
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

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

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