Test Index

Circles Part 2

Section: Mathematics

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Question : 77 of 87

Marks: +1, -0

Let the tangent to the circle

C_{1}: x^{2}+y^{2}=2

M(-1,1)

intersect the circle

C_{2}:(x-3)^{2}+(y-2)^{2}=5

, at two distinct points

A

B

. If the tangents to

C_{2}

A

B

N

, then the area of the triangle

A N B

[29-Jun-2022-Shift-1]

$\frac{1}{2}$
$\frac{2}{3}$
$\frac{1}{6}$
$\frac{5}{3}$

Solution: 👈: Video Solution

Equation of tangent at point

M

is

T=0

\Rightarrow x x_{1}+y y_{1}=2

\Rightarrow -x+y=2

\Rightarrow y=x+2

Putting this value to equation of circle

C_{2}

,

(x-3)^{2}+(y-2)^{2}=5

\Rightarrow (x-3)^{2}+x^{2}=5

\Rightarrow x^{2}-6x+9+x^{2}=5

\Rightarrow 2x^{2}-6x+4=0

\Rightarrow x^{2}-3x+2=0

\Rightarrow (x-2)(x-1)=0

\Rightarrow x=1,2

when

x=1, y=3

and when

x=2, y=4

\therefore

Point

A(1,3)

and

B(2,4)

Now, equation of tangent at

A(1,3)

on circle

(x-3)^{2}+(y-2)^{2}=5

or

x^{2}+y^{2}-6x-4y+8=0

is

T=0

x x_{1}+y y_{1}+g(x+x_{1})+f(y+y_{1})+C=0

\Rightarrow x+3y-3(x+1)-2(y+3)+8=0

\Rightarrow x+3y-3x-3-2y-6+8=0

\Rightarrow -2x+y-1=0

\Rightarrow 2x-y+1=0

Similarly tangent at

B(2,4)

is

2x+4y-3(x+2)-2(y+4)+8=0

\Rightarrow 2x+4y-3x-6-2y-8+8=0

\Rightarrow -x+2y-6=0

\Rightarrow x-2y+6=0

Solving equation (1) and (2), we get

x-2(2x+1)+6=0

\Rightarrow x-4x-2+6=0

\Rightarrow -3x+4=0

\Rightarrow x=\frac{4}{3}

\therefore y=2\times\frac{4}{3}+1=\frac{11}{3}

\therefore \text{Point } N=\left(\frac{4}{3},\frac{11}{3}\right)

Now area of the triangle ANB

=\frac{1}{2}\begin{vmatrix} x_{1} & y_{1} & 1 \\ x_{2} & y_{2} & 1 \\ x_{3} & y_{3} & 1 \end{vmatrix}

=\frac{1}{2}\begin{vmatrix} 1 & 3 & 1 \\ 2 & 4 & 1 \\ \frac{4}{3} & \frac{11}{3} & 1 \end{vmatrix}

=\frac{1}{2}\left[1\left(4-\frac{11}{3}\right)-3\left(2-\frac{4}{3}\right)+1\left(\frac{22}{3}-\frac{16}{3}\right)\right]

=\frac{1}{2}\left[\frac{1}{3}-\frac{6}{3}+\frac{6}{3}\right]

=\frac{1}{6}

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