Analytic Geometry

Question

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 $ ?

Answer 1

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.

Answer 2

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

Answer 3

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 $$ \Large\color{blue}{(x+2)^2+(y-5)^2=10}. $$

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