Test Index

Circle

Section: Mathematics

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Question : 38 of 89

Marks: +1, -0

The angle between the tangents drawn from the point

(2,2)

x^2+y^2+4 x+4 y+c=0

\cos^{-1} \left( \frac{7}{16} \right)

. If two such circles exist, then sum of the values of c is

[AP EAPCET 21 May 2025 S2]

$16$
$20$
$-20$
$-16$

Solution:

We have the circle

x^2+y^2+4 x+4 y+c=0

C=\text{Centre}=(-2,-2)

A C=\text{radius}=\sqrt{8-C}

Angle between tangents

\cos 2\theta = \frac{7}{16}, CP = \sqrt{16+16}=4\sqrt{2}

\because \sin \theta = \frac{3}{4\sqrt{2}}

Using

\sin \theta = \frac{r}{CP} = \frac{\sqrt{8-C}}{4\sqrt{2}} = \frac{3}{4\sqrt{2}}

\because 8-C=9

C=-1

Now, another circle is possible when angles be supplementary

\because \theta

is changed by

\pi - 2\theta

\because \sin \left( \frac{\pi}{2} - \theta \right) = \cos \theta

\because \cos \theta = \sqrt{\frac{1+\cos 2\theta}{2}} = \frac{\sqrt{23}}{4\sqrt{2}}

\because \cos \theta = \frac{\sqrt{8-C}}{4\sqrt{2}} = \frac{\sqrt{23}}{4\sqrt{2}}

\because 8-c=23

-c=-15

\because

Now, sum of all possible values of

C

.

=-1-15=-16

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