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  • The circles <img alt="x^2+y^2-4 x-81=0, x^2+y^2+24 x-81=0" src="https://entrancecorner.oncodecogs.com/gif.latex?x%5E2&plus;y%5E2-4%20x-81%3D0%2C%20x%5E2&plus;y%5E2&plus;24%20x-81%3

The circles x^2+y^2-4 x-81=0, x^2+y^2+24 x-81=0 intersect each other at points A and B. A line through point A meet one circle at  P and a parallel line through B meet the other circle at Q. Then the locus of the mid point of PQ is

Option: 1

(x+5)^2+(y+0)^2=25 \\


Option: 2

(x-5)^2+(y-0)^2=25 \\


Option: 3

x^2+y^2+10 x=0 \\


Option: 4

x^2+y^2-10 x=0


Answers (1)

best_answer

\\\text{Solving }\: \: x^2+y^2-4 x-81=0\\ \\\text{and } x^2+y^2+24 x-81=0\\ \\\text{we get} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: x=0\\ \\\text{and }\: \: \: \: \: \: \: \: \: \: \: \:\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: y= \pm 9

\therefore A(0,-9) \text { and } B(0,9)

Equation of line through A(0,-9) is 

y+9=m x \Rightarrow y=m x-9\: \: \: \: \: \: \: \: \: \: \: \: ....(iii)

Solving Eqs. (i) and (iii), then 

\begin{aligned} & x^2+(m x-9)^2-4 x-81=0 \\ \\& x^2\left(1+m^2\right)-(18 m+4) x=0 \\ \\\therefore \quad & p\left(\frac{18 m+4}{1+m^2}, \frac{9 m^2+4 m-9}{1+m^2},\right) \end{aligned}

and equation of line through B(0,9) and parallal to (iii) is 

\\y=m x+\lambda \Rightarrow 9=0+\lambda \Rightarrow y=m x+9\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ....(iv)\\ \\\text{Solving Eqs. (ii) and (iv), then}\\ \begin{aligned} \\& \Rightarrow \quad x^2\left(1+m^2\right)+(18 m+24) \cdot x=0 \\ \\& \therefore Q\left(-\frac{(18 m+24)}{1+m^2}, \frac{-9 m^2-24 m+9}{1+m^2},\right) \end{aligned}

Let mid point of PQ is (h,k), then 

\begin{aligned} & \quad 2 h=-\frac{20}{1+m^2} \Rightarrow h=-\frac{10}{\left(1+m^2\right)}\: \: \: \: \: \: \: \: \: \: \: \: \: ...(v) \\ \\& \text { and } 2 k=-\frac{20 m}{1+m^2} \Rightarrow k=-\frac{10 m}{1+m^2}\: \: \: \: \: \: \: \: \: \: \: \: \: \: .....(vi) \end{aligned}

\\\text{From Eqs. (v) and (vi)}, \frac{k}{h}=m \\\\\text{Substituting the value of m in (v)}

\begin{aligned} & \text { Then } h\left(1+\frac{k^2}{h^2}\right)=-10 \\ \\& \Rightarrow \quad h^2+k^2=-10 h \\ \\& \Rightarrow h^2+k^2+10 h=0\\ \end{aligned} \\\therefore \text{Locus of mid point of PQ is}\\ or \begin{gathered} \\x^2+y^2+10 x=0 \\ (x+5)^2+(y+0)^2=25 \end{gathered}

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