Let $S_1:x^2+y^2+4y-1=0, S_2: x^2+y^2+6x+y+8=0; S_3: x^2+y^2-4x-4y-37=0$ touch each other. Let $P_1,P_2,P_3$ be their contact points and let $C_1,C_2,C_3$ be their centers. Find $\frac{\Delta P_1P_2P_3}{\Delta C_1C_2C_3}$
I can find the points P using radical axes formula, but this process is quite lengthy. Is there any property I can exploit for this question ?
On
Is there any property I can exploit for this question ?
Presumably the circles are tangential (otherwise there would be more than $6$ points). So the points $P_i$ will be be colinear with the centers and you can figure $P_i$ from that. That should shave minutes off your calculations.
$S_1:x^2+y^2+4y-1=0::x^2 + (y+2)^2 = 5$ Center is $(0,-2)$ and radius if $\sqrt 5$.
$S_2: x^2+y^2+6x+y+8=0:: (x+3)^2 + (y+\frac 12)^2=\frac 54$; Center is $(-3,-\frac 12)$ and radius is $\frac {\sqrt 5}2$.
Verify the person who gave to problem was telling the truth and that these circles are tangential. $||(0,-2),(-3,-\frac 12)|| = \sqrt{3^2 + (-2+\frac 12)^2}=\sqrt{9+\frac {9}{4}}=\frac {3\sqrt 5}2= r_1 + r_2$ so yes they are tangential and $P_1 = C_1 + \frac {r_1}{r+1 + r_2}[(C_2 - C_1)] = (0,-2) + \frac 23(-3,-\frac 12+2)=(-2,-1)$
And so on.
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The radii are in the ratios $r_1: r_2: r_3 =1:2:6$, implying right $\triangle C_1C_2C_3$ with $\sin C_3 = \frac35$ and $\cos C_3 = \frac45$, as pointed out by @MathLover. Then
\begin{align} \frac {\triangle P_1P_2P_3}{ \triangle C_1C_2C_3} = & \frac {\triangle P_1P_2C_3 - \triangle P_1P_3C_3 - \triangle P_2P_3C_3}{ \triangle C_1C_2C_3}\\ = & \frac {\frac12 r_3^2 \sin C_3- \frac12 r_1r_3 \cos C_3 - \frac12 r_2r_3}{ \frac12 (r_3-r_2)(r_1+r_2)}\\ = & \frac {\frac35- \frac{r_1}{r_3} \frac45 - \frac{r_2}{r_3}}{ (1-\frac{r_2}{r_3})(\frac{r_1}{r_3} +\frac{r_2}{r_3})}=\frac25\\ \end{align}
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I can find the points P using radical axes formula, but this process is quite lengthy.
The tangency point is collinear with two centers and divides the join of centers in ratio of radii.
The catch about this problem is realizing that $S_1,S_2$ are internally tangent to $S_3$ which can be done by checking the powers of $O_1,O_2$ wrt $S_3$, the one with suspiciously large radius.
Rest of the problem is quite easy. I calculated the $P$s using section formula as in following doodle in my notebook.
These come as nice integers. Thus $$\frac{[P_1P_2P_3]}{[C_1C_2C_3]}=\frac{\begin{vmatrix} -2 & -1 & 1 \\ -4 & -1 & 1 \\ -1 & -4 & 1 \end{vmatrix} }{ \begin{vmatrix} 0 & -2 & 1 \\ -3 & -1/2 & 1 \\ 2 & 2 & 1 \end{vmatrix}}$$
On $R_2 \to R_2-R_1$ and $R_3 \to R_3-R_1$, $$=\frac{\begin{vmatrix} -2 & 0 \\ 1 & -3 \end{vmatrix} }{ \begin{vmatrix} -3 & 3/2 \\ 2 & 4 \end{vmatrix}}$$
Hence $$\frac{[P_1P_2P_3]}{[C_1C_2C_3]}=\frac{2}{5}$$
You do not necessarily have to find the points of contacts to find the area of triangles here.
$\small S1: x^2+(y+2)^2 = 5, S2: (x+3)^2 + (y + \frac{1}{2})^2 = \frac{5}{4}, S3: (x-2)^2+(y-2)^2 = 45$
Based on their centers and radii, we observe that centers $\small C_1, C_2$ are inside $\small S_3$ so if $\small S_1, S_2$ are tangent to $S_3$, they must be tangent to it on the inside of $\small S_3$. Also $\small S_1, S_2$ seem to be tangent externally as $\small C_1 C_2 = r_1 + r_2$. Here is a diagram that depicts the scenario.
$\displaystyle \small C_1C_2 = r_1 + r_2 = \frac{3\sqrt5}{2}, C_2C_3 = r_3 - r_2 = \frac{5\sqrt5}{2}, C_3C_1 = r_3 - r_1 = 2\sqrt5$
Now please see that $\small (C_1C_2)^2 + (C_3C_1)^2 = (C_2C_3)^2$ so $\small \angle C_2C_1C_3 = 90^0$ and $\small P_2P_3 = r_1 \sqrt2 = \sqrt{10}$
$\displaystyle \small \cos \angle C_1C_2C_3 = \frac{C_1C_2}{C_2C_3} = \frac{3}{5} \implies \cos \angle P_1C_2P_3 = - \frac{3}{5}$
Using law of cosine in $\small \triangle P_1C_2P_3, (P_1P_3)^2 = 2 r_2^2(1 - \cos \angle P_1C_2P_3) \implies P_1P_3 = 2$
Similarly, $\small (P_1P_2)^2 = 2r_3^2 (1 - \cos \angle C_1C_3C_2) \implies P_1P_2 = 3 \sqrt2$
Now we know all sides of $\small \triangle P_1P_2P_3$ and can find its area. Also, $\small \triangle C_1C_2C_3$ is a right angled triangle with known sides.