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Lets say the circle touches $3x + 4y = 0$ and $3x + 4y -50 = 0$, and passes through the point $(2,1)$.

My approach was to let the centre of the circle be $(x,y)$, and set up $3$ equations:

$$\begin{align}&(2-x)^2 + (1-y)^2 = r^2\\&r = \frac{3x+4y}{\sqrt{25}}\\&r=\frac{3x+4y-50}{\sqrt{25}}\end{align}$$

where r is the radius. However the problem is when I consider the 2nd and 3rd equations, I would get $0 = -50$ which does not make sense. Where did I make a mistake?

user773458
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gatiskandis
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2 Answers2

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Actually you have a problem with the second and the third equations. The distance of a generic point $(x_0,y_0)$ from a line $ax+by+c=0$ is: $$d=\frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}$$ There's an absolute value! So, if $(x_c,y_c)$ is the center of the circle, your equations should be: $$\begin{align}&r = \frac{|3x_c+4y_c|}{\sqrt{25}}\\&r=\frac{|3x_c+4y_c-50|}{\sqrt{25}}\end{align}\:\Longrightarrow\:y_c=-\frac{3}{4}x_c+\frac{25}{4}$$ This implies that: $$r=\frac{|3x_c+4y_c|}{5}=\frac{|3x_c+4\left(-\frac{3}{4}x_c+\frac{25}{4}\right)|}{5}=\frac{|3x_c-3x_c+25|}{5}=5$$ The $x$-coordinate $x_c$ of your center satisfies: $$25=(2-x_c)^2+(1-y_c)^2$$ which can be simplified to: $$5x_c^2-38x_c+21=0\:\Longrightarrow\:x_c=\frac{3}{5},7$$ So there are two circles fulfilling your conditions: $$\left(x-\frac{3}{5}\right)^2+\left(y-\frac{29}{5}\right)^2=25,\quad\left(x-7\right)^2+\left(y-1\right)^2=25$$

Further details: The equation $y_c=-\frac{3}{4}x_c+\frac{25}{4}$ can be derived straightforwardly. Since we know that the circle is tangent to the parallel lines $\ell_1:3x+4y=0$ and $\ell_2:3x+4y-50=0$, the center must lie on the line $\ell_3$ parallel to $\ell_1$ and $\ell_2$ such that $d(\ell_1,\ell_3)=d(\ell_2,\ell_3)$. This latter fact directly tells us that: $$3x_c+4y_c-25=0$$ as wanted. Moreover, the radius is equal to: $$r=d(\ell_1,\ell_3)=\frac{|0-25|}{\sqrt{3^2+4^2}}=5$$

user773458
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  1. The distance between the 2 lines is given by $\pm {\dfrac {C_1 – C_2}{\sqrt {A^2 + B^2}}}$ = ... = 10.

  2. The radius of the required circle is therefore = …. = 5.

  3. Let X(h, k) be the center of the required circle (called Q). Note that X should lie on the line …. $L: 3x + 4y – 25 = 0$.

  4. Let P = (2, 1). Then the QP = 5. Get the equation of the circle K (centered at P with radius = 5).

  5. Solve L and K to get TWO sets of values for (h, k).

  6. Use h, k , 5 to build the equations of the required circles.

Mick
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