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Finding the equation of the circle which touches the pair of lines $7x^2 - 18xy +7 y^2 = 0$ and the circle $x^2 + y^2 - 8x -8y = 0$ and contained in the given circle??

My attempt The centre of required circle would lie on angle bisector of the pair of lines ie $x=y$.

Assuming circle to be $(x-h)^2+(y-h)^2=r^2$

Now $2(h-8)^2=r^2$ ( distance between the extreme of larger circle and center of contained circle,)

I am unable to frame a second equation . One way would be to calculate the angle between pair of straight lines and use it to find a relation between $r$ and $h$.

However I was looking for a better solution or suggestion ?

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  • $\begingroup$ Going through your past question I notice that nobody has taken the time yet to tell you that we MathJax on this site to format our maths. You can find a basic tutorial here. Please have a look at it so you will be able to format your own posts. $\endgroup$ – gebruiker Oct 6 '17 at 17:01
  • $\begingroup$ For the center I got $y=x$ or $y=-x$. $\endgroup$ – Michael Rozenberg Oct 6 '17 at 17:11
  • $\begingroup$ @MichaelRozenberg forgot to mention that circle lies in first quadrant. $\endgroup$ – Seema Prasad Oct 6 '17 at 17:30
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You are almost there. All you need is to use the fact that the circle center is equidistant from both lines. In particular, using the distance formula, you can write $$r^2 = \frac{((9+4\sqrt{2})h-7h)^2}{(9+4\sqrt{2})^2+7^2} = 2(h-8)^2 \implies h=6,12.$$ Since the center $(h,h)$ lies in the bigger circle, $h\neq 12$. Consequently, $h=6$ and $r^2=8.$ So $$(x-6)^2+(y-6)^2=8$$ is the equation of the sought circle.

Note that the tangent lines are $7y = (9\pm 4\sqrt{2})x.$

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  • $\begingroup$ Thanks . I was trying to avoid calculating the lines. $\endgroup$ – Seema Prasad Oct 6 '17 at 17:30
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Hint:

enter image description here

The sought circle is the incircle of $\triangle ABC$.

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The angle bisector of the two lines $y=\dfrac{1}{7} \left(9 \pm4 \sqrt{2} \right)x$ is the line $y=x$

The wanted circle has then centre $H(k,k)$ and its radius is the distance from the given lines $\left(9+4 \sqrt{2}\right) x-7 y=0$

$$r_k=\frac{\left|\left(9+4 \sqrt{2}\right) k-7 k\right|}{\sqrt{\left(9+4 \sqrt{2}\right)^2+49}}=\frac{\left|2 \left(1+2 \sqrt{2}\right) k\right|}{12+3 \sqrt{2}}$$ The wanted circle must also be tangent to the given circle $x^2-8 x+y^2-8 y=0$ having centre $C(4,4)$ and radius $R=4\sqrt{2}$

A circle is internally tangent to another when the distance of the centres is equal to the difference of the radii (in absolute value)

Therefore we must have $CH=R-r_k$

that is $$\sqrt{(4-k)^2+(4-k)^2}=4\sqrt{2}-\frac{\left|2 \left(1+2 \sqrt{2}\right) k\right|}{12+3 \sqrt{2}}$$ which simplified, noticing that $k>4$ becomes $$\sqrt{2}(k-4)=\frac{2 \left(1+2 \sqrt{2}\right) (12-k)}{3 \left(4+\sqrt{2}\right)}$$ After a look to the conditions we can say that $k>4$ so the previous equation becomes $$\left(12+3 \sqrt{2}\right) \sqrt{2} (k-4)=2 \left(1+2 \sqrt{2}\right) k$$

Solution is $k=6$ and the wanted circle has equation

$(x-6)^2+(y-6)^2=8\to \color{red}{x^2+y^2-12x-12y+64=0}$

enter image description here

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