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Circle

Section: Mathematics

Question : 88 of 89

Marks: +1, -0

2 x-3 y+3=0

x+2 y+k=0

S \equiv x^2+y^2+8 x-6 y-24=0

\left(\frac{k}{4}, \frac{k}{3}\right)

S=0

Solution:

Let (

a, b

be the point on the line

x+2 y+K=0

We have, equation of circle is

x^2+y^2+8 x-6 y-24=0

Equation of polar for the given circle is obtained as

\; a x+b y+4(x+d)-3(y+b)-24=0

\;(a+4 x+(b-3) y+4 a-3-24

Now, comparing this line with

2 x-3 y+3=0

we get

\;\frac{a+4}{2}=\frac{b-3}{-3}=\frac{4 a-y-24}{3}

\; -3 a-12=2 b-6 \Rightarrow 3 a+2 b=-6

\;\;\text{ and }\; 3 a+12=8 a-6 b-4 s

\; \Rightarrow 5 a-6 b=60 \dots \;\text{ (ii) }\;

On solving Eqs. (i) and (ii), we get

a=3 \;\text{ and }\; b=\;\frac{-15}{2}

Now.

\left(3, \frac{15}{2}\right)

lies on line

x+2 y+k=0

3-15+K=0 \Rightarrow K=12

Required length of tangent from a point

\; \left(\;\frac{12}{4},\;\frac{12}{3}\right) \;\text{ i.e. }\;(3,4)

\; =\sqrt{3^3+4^2+24-24-24}

\; =\sqrt{9+16-24}=1

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