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A circle has equation, $x^2+y^2+14x+4y-19=0$

A smaller circle of centre $C$ shares a common tangent $y=3-x$ at the point $P$

The radius of the larger circle is three times the radius of the smaller circle.

Find the equation of the smaller circle.

visual of question

I've spent a while playing around with this question. I have manged to solve it through the use of a diagram but I cannot see a more... Mathematical solution.

Through "counting boxes" I found the equation to be $(x-1)^2+(y-6)^2=(2\sqrt2 )^2$

Any input is much appreciated. Thanks in advance.

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Here is a "mathematical" way in which we do not need to draw a diagram:

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Suppose the equation of the small circle is $$(x-a)^2+(y-b)^2=(2\sqrt2)^2$$ Then differentiate both sides w.r.t $\,x$, and we have $$2(x-a)+2(y-b)\frac{dy}{dx}=0$$ Since the tangent point is $(-1,4)$ and the the slope of the tangent line is $-1$, we have $$2(-1-a)+2(4-b)(-1)=0$$ $$\Rightarrow\ \ b=5+a$$ Back to the equation of the small circle, $$(-1-a)^2+(4-(a+5))^2=8$$ $$\Rightarrow\quad 2(a+1)^2=8\quad$$ $$\Rightarrow\quad a=-3,1\ \ \ \&\ \ \ b=2,6$$ Thus, the equation of the small circle is: $$(x+3)^2+(y-2)^2=8$$ $$\text{or}\quad\ (x-1)^2+(y-6)^2=8\qquad$$

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Sketch: First of all find the coordinates of $P$. Then write the equation of the line $s$ perpendicular to $y=3-x$ which passes through $P$. Then write a generic point on $s$ and find the ones whose distance from $P$ is equal to $R/3$, where $R$ is the radius of the big circle (You will find two points: one within the big circle and one outside. Of course you need to select the latter).

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    $\begingroup$ I don't think there's a compelling reason to select one point over the other just based on the given question. $\endgroup$ – Chappers Mar 2 '17 at 21:37
  • $\begingroup$ You are perfectly right! Then the problem doesn't have a unique answer, which is fine! $\endgroup$ – Stefano Mar 2 '17 at 23:35
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Hint: The center of large circle is $C(-7,-2)$. Obtain a line perpendicular to $L:\,y=3-x$ which pass through $C$, named $L'$, Find the distance $C$ from $L$ and with $\frac13$ you find redius of smaller circle. find it's center on line $L'$ to finish your work. The contact point $P:\,LL'$ help more because if $P(\alpha,\beta)$ and $C'(a,b)$ then $$\alpha=\dfrac{-7\times3+a\times1}{4},\beta=\dfrac{-2\times3+b\times1}{4}$$

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