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  • The radical axis of the circles  <img alt="\mathrm{x^2+y^2+2 g x+2 f y+c=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7Bx%5E2&plus;y%5E2&plus;2%20g%20x&plus;2

The radical axis of the circles  \mathrm{x^2+y^2+2 g x+2 f y+c=0} and  \mathrm{2x^2+2y^2+3 x+8 y+2c=0} touches the circle \mathrm{x^2+y^2+2 x+2 y+1=0} ,then f is equal to __________ 

Option: 1

 

2


Option: 2

3


Option: 3

4


Option: 4

5


Answers (1)

best_answer

The second circle can be reduced to standard form on
dividing by 2 so that its equation is

\mathrm{S_2=x^2+y^2+\frac{3}{2} x+4 y+c=0}

The radical axis of circles \mathrm{S_1=0 \text { and } S_2=0} is given by

\mathrm{\begin{aligned} & S_1-S_2=0 \\ & \Rightarrow 2 x(g-3 / 4)+2 y(f-2)=0 \\ & \Rightarrow x(g-3 / 4)+y(f-2)=0 \end{aligned}}

It touches the circle S_3=0  whose centre is (–1, –1) and
radius 1. The condition of tangency i.e. p = r gives

\mathrm{\left|\frac{-1 \cdot(g-3 / 4)-(f-2) 1}{\sqrt{\left|(g-3 / 4)^2+(f-2)^2\right|}}\right|=1}

Squaring we get

\mathrm{ \begin{aligned} & (g-3 / 4)^2+(f-2)^2+2(g-3 / 4)(f-2)=(g-3 / 4)^2+(f-2)^2 \\ & \therefore \quad 2(g-3 / 4)(f-2)=0 . \end{aligned}}

\mathrm{Hence \; either \; g-3 / 4 \; or \; f-2=0}
\mathrm{\therefore \quad \; Either\; g=3 / 4 \; or\; f=2.}

 

 

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Nehul

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