Find-the-equation-of-the-circle-which-touches-the-circles-x-y-2x-4y-1-x-y-12x-2y-4-and-x-y-2x-12y-12-0-

Question Number 224360 by Tawa11 last updated on 05/Sep/25

Find the equation of the circle which touches
the circles x² + y² – 2x + 4y = 1, x² + y² – 12x + 2y = 4
and x² + y² + 2x – 12y + 12 = 0.

Commented by TonyCWX last updated on 05/Sep/25

$${Exactly}\:{the}\:{same}\:{question}\:{in}\:{my}\:{WhatsApp}\:{group}. \\ $$

Answered by TonyCWX last updated on 05/Sep/25

Commented by Ghisom_ last updated on 05/Sep/25

$$\mathrm{there}\:\mathrm{should}\:\mathrm{be}\:\mathrm{4}\:\mathrm{circles}… \\ $$

Commented by TonyCWX last updated on 05/Sep/25

Commented by Tawa11 last updated on 05/Sep/25

$$\mathrm{Thanks}\:\mathrm{sir},\:\mathrm{I}\:\mathrm{appreciate}. \\ $$

Commented by TonyCWX last updated on 05/Sep/25

$${What}\:{do}\:{you}\:{mean}? \\ $$

Commented by Ghisom_ last updated on 05/Sep/25

in these formulas (a−α)^2 +(b−β)^2 =(γ+r)^2 r is orientated. if you look at your picture you can see that “+r, +r, +r” gives 2 circles touching the 3 given ones from the outside “+r, −r, +r” gives 2 circles inside the blue one rounded to 3 digits: ((x_(center) ),(y_(center) ),(r) )_ ∈{ (((−.849)),((.412)),((.590)) ) , (((13.8)),((10.8)),((20.6)) ) , (((.256)),((.749)),((.399)) ) , (((6.72)),((1.19)),((4.10)) ) }

$$\mathrm{in}\:\mathrm{these}\:\mathrm{formulas} \\ $$$$\left({a}−\alpha\right)^{\mathrm{2}} +\left({b}−\beta\right)^{\mathrm{2}} =\left(\gamma+{r}\right)^{\mathrm{2}} \\ $$$${r}\:\mathrm{is}\:\mathrm{orientated}.\:\mathrm{if}\:\mathrm{you}\:\mathrm{look}\:\mathrm{at}\:\mathrm{your}\:\mathrm{picture} \\ $$$$\mathrm{you}\:\mathrm{can}\:\mathrm{see}\:\mathrm{that} \\ $$$$“+{r},\:+{r},\:+{r}''\:\mathrm{gives}\:\mathrm{2}\:\mathrm{circles}\:\mathrm{touching}\:\mathrm{the} \\ $$$$\mathrm{3}\:\mathrm{given}\:\mathrm{ones}\:\mathrm{from}\:\mathrm{the}\:\mathrm{outside} \\ $$$$“+{r},\:−{r},\:+{r}''\:\mathrm{gives}\:\mathrm{2}\:\mathrm{circles}\:\mathrm{inside}\:\mathrm{the} \\ $$$$\mathrm{blue}\:\mathrm{one} \\ $$$$\mathrm{rounded}\:\mathrm{to}\:\mathrm{3}\:\mathrm{digits}: \\ $$$$\begin{pmatrix}{{x}_{\mathrm{center}} }\\{{y}_{\mathrm{center}} }\\{{r}}\end{pmatrix}_{} \:\in\left\{\begin{pmatrix}{−.\mathrm{849}}\\{.\mathrm{412}}\\{.\mathrm{590}}\end{pmatrix}\:,\:\begin{pmatrix}{\mathrm{13}.\mathrm{8}}\\{\mathrm{10}.\mathrm{8}}\\{\mathrm{20}.\mathrm{6}}\end{pmatrix}\:,\:\begin{pmatrix}{.\mathrm{256}}\\{.\mathrm{749}}\\{.\mathrm{399}}\end{pmatrix}\:,\:\begin{pmatrix}{\mathrm{6}.\mathrm{72}}\\{\mathrm{1}.\mathrm{19}}\\{\mathrm{4}.\mathrm{10}}\end{pmatrix}\:\right\} \\ $$

Commented by TonyCWX last updated on 05/Sep/25

$${Rounded}\:{to}\:\mathrm{3}\:{s}.{f}. \\ $$

Commented by Ghisom_ last updated on 26/Sep/25

your quadratic has 2 solutions, both are correct, even though the 2^(nd) one is <0 as I wrote r is orientated just try it, you′ll see it

$$\mathrm{your}\:\mathrm{quadratic}\:\mathrm{has}\:\mathrm{2}\:\mathrm{solutions},\:\mathrm{both}\:\mathrm{are} \\ $$$$\mathrm{correct},\:\mathrm{even}\:\mathrm{though}\:\mathrm{the}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{one}\:\mathrm{is}\:<\mathrm{0} \\ $$$$\mathrm{as}\:\mathrm{I}\:\mathrm{wrote}\:{r}\:\mathrm{is}\:\mathrm{orientated} \\ $$$$\mathrm{just}\:\mathrm{try}\:\mathrm{it},\:\mathrm{you}'\mathrm{ll}\:\mathrm{see}\:\mathrm{it} \\ $$

Commented by TonyCWX last updated on 05/Sep/25

$${I}\:{have}\:{seen}\:{them}. \\ $$$${Thanks}. \\ $$

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