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How does one solve:

Find the equation of the circle which has it's center on the line $y= 3-x$ , and which has as tangents the lines $ 2y-x = 22, $ $ 2x+y=11 $ ?

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Let $(a,b)$ be the center of the circle. Since the center lies on the line $y=3-x$, then we have $b=3-a$. The equation of the circle if the center on $(a,b)$ and radius $r$ is $$ (x-a)^2+(y-b)^2=r^2\tag1 $$ and the equation of its tangent line is $$ (x_c-a)(x-a)+(y_c-b)(y-b)=r^2,\tag2 $$ where $(x_c,y_c)$ is point of contact. Equation $(2)$ can be written as $$ (x_c-a)x+(y_c-b)y=r^2+a(x_c-a)+b(y_c-b).\tag3 $$ The tangent lines are $$ -x+2y=22\tag4 $$ and $$ 2x+y=11.\tag5 $$ Using $b=3-a$, comparing $(3)$ and $(4)$ yields $x_1-a=-1$, $y_1-b=2$, and $$ \begin{align} r^2+a(x_1-a)+b(y_1-b)&=22\\ r^2+a(-1)+b(2)&=22\\ r^2-a+2(3-a)&=22\\ r^2-3a&=16.\tag6 \end{align} $$ Similarly, comparing $(3)$ and $(5)$ yields $x_2-a=2$, $y_2-b=1$, and $$ \begin{align} r^2+a(x_2-a)+b(y_2-b)&=11\\ r^2+a&=8.\tag7 \end{align} $$ Solving $(6)$ and $(7)$ yields $a=-2$, $b=5$, and $r^2=10$. Thus, using $(1)$, the equation of the circle is $$ \Large\color{blue}{(x+2)^2+(y-5)^2=10}. $$


$$\Large\color{blue}{\text{# }\mathbb{Q.E.D.}\text{ #}}$$

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not bad, there are two answers however... – Bak1139 Apr 21 '14 at 10:09

Hint: Can you use the two tangent lines you have to find a line which must pass through the centre of the circle?

I suggest sketching a diagram.

And I suggest you give more detail of what you know about tangents and circles, and what you have tried.

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thing is the radius is unknown nor the center, so even when such an equation produced, it doesn't help much...also they have two possible equations as answers which is quite strange... – Bak1139 Apr 21 '14 at 8:58
@Bak1139 You have one line which passes through the centre of the circle already. If you know another one (different and not parallel) you will find that they intersect at the centre without having to compute the radius first. If you draw a diagram and consider how to construct such a thing you should see that there are two possibilities. – Mark Bennet Apr 21 '14 at 9:05
perhaps 2, but then , why not 4? – Bak1139 Apr 21 '14 at 9:09
@Bak1139 Draw a diagram and you will see it. – Mark Bennet Apr 21 '14 at 9:15

The sphere center must lie on the dichotomus of the two lines which isn't hard to find. That fact together with the other line equation will give you the solution.QED

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that's a good point.... – Bak1139 Apr 21 '14 at 8:59

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