# Coordinates of the center of the circle

I am stuck on this problem:

If the lines $y=x+\sqrt{2}$ and $y=x-2\sqrt{2}$ are two tangents of a circle and $(0,\sqrt{2})$ lies on this circle then what is the equation of the circle?

I found out the distance between the two tangents $y=x+\sqrt{2}$ and $y=x-2\sqrt{2}$ is $3$. The radius is $3/2$ but I don't know how to find the center. I tried forming the equations but could not succeed. Please tell me the easiest way. Thank you

• Draw a sketch with those two lines. What's the equation of the line on which the center must lie? Now, given that line, the radius and a point, you should be able to narrow it down. – Simon S May 27 '15 at 12:51

Hint: A sketch of the situation should be helpful:

What can we say about the position of the center of a circle with those two tangents?

The equation of a circle with radious $r$ and center $(x_0,y_0)$ is: $$(x - x_0)^2 + (y - y_0)^2 = r^2$$

Solution:

a) Center line: The center points lie on a line in the middle of the lines, thus on \begin{align} y_c(x) &= \frac{1}{2}(y_1(x) + y_2(x)) \\ &= \frac{1}{2}((x + \sqrt{2}) + (x - 2 \sqrt{2})) \\ &= x - \frac{1}{\sqrt{2}} \end{align}

b) Radius: The radius of the circle is two times the distance of the tangent lines.

An orthogonal line to both tangents is $y = - x$ it intersects $y_1$ if $$x + \sqrt{2} = - x \iff 2x = -\sqrt{2} \iff x = - 1/\sqrt{2}$$ thus at $P = (-1/\sqrt{2}, 1/\sqrt{2})$. The distance to the origin is $\lVert OP \rVert = \sqrt{1/2 + 1/2} = 1$

It intersects $y_2$ if $$x - 2 \sqrt{2} = -x \iff 2x = 2 \sqrt{2} \iff x = \sqrt{2}$$ thus at $Q = (\sqrt{2}, -\sqrt{2})$. The distance to the origin is $\lVert OQ \rVert = \sqrt{2 + 2} = 2$. So both tangents are at a distance $3$ and the radius is $3/2$.

c) Equation of any circle with those tangents:

So all circles with those two tangents have the equation $$(x - x_0)^2 + \left(y - x_0 + \frac{1}{\sqrt{2}}\right)^2 = 9/4$$

d) The circle with those tangents that contains $(0, \sqrt{2})$:

We want the one that contains $(0, \sqrt{2})$ thus $$9/4 = x_0^2 + (\sqrt{2} - x_0 + 1/\sqrt{2})^2 = x_0^2 + (3/\sqrt{2} - x_0)^2 = x_0^2 + 9/2 + x_0^2 - 3\sqrt{2} x_0 \Rightarrow \\ -9/4 = 2 x_0^2 - 3\sqrt{2} x_0 \Rightarrow \\ -9/8 = x_0^2 - (3/\sqrt{2}) x_0 = (x_0-3/(2\sqrt{2}))^2 - 9/8 \Rightarrow \\ x_0 = \frac{3}{2\sqrt{2}}$$ This gives the center $C=(3/(2\sqrt{2}), 3/(2\sqrt{2}) - 1/\sqrt{2}) = (3/(2\sqrt{2}), 1/(2\sqrt{2}))$. And we have the equation $$\left(x - \frac{3}{2\sqrt{2}} \right)^2 + \left(y - \frac{1}{2\sqrt{2}} \right)^2 = 9/4$$

• Yes, exactly what I was trying to suggest – Simon S May 27 '15 at 13:03
• Thank you ! I know all these things and I also understood the equation by @simon but I don't know how am I going to get the coordinates by that. I am sorry .I am very weak in math :( – Seeker1201 May 27 '15 at 13:37
• Move the circle in this diagram down until it is in place. Draw in the radius between $(0,\sqrt 2)$ and the center. What is the equation of that straight line? As it is perpendicular to $y = x + \sqrt 2$, the equation of the new straight line must have slope $-1/1$, i.e., $y = - x + c$ for some constant $c$. Well, when $x = 0, y = \sqrt 2$, so in fact the equation of the new line segment is $$y = -x + \sqrt 2$$ The equation of the line of centers is $y = x + d$ for some $d$. Solve for $d$. Then you have two equations to solve where the lines intersect; that's where the center is. – Simon S May 27 '15 at 14:36
• What is the software used to make such diagrams? – Aadam Apr 6 '16 at 17:07
• That is GeoGebra. It has nice 2D and 3D modes and is available on many plattforms, like Windows, Linux and iOS for iPad, there is also a version that runs as web browser applet and a service to host interactive graphics on the web (GeoGebra Tube). It is used in school as well. While the main interfaces are click and point editors and the command line it is also possible to program complex geometric scenes by its own script language or in JavaScript. – mvw Apr 6 '16 at 19:13

hints:

Since $\;(0,\,\sqrt2)\;$ is both on the circle and on one of the lines (why and which line?), the line must be perpendicular to the radius of the circle at this point (why?), so the circle's center is on the line

$$y-\sqrt2=-(x-0)\implies y=-x+\sqrt2$$

Since the given second line is parallel to the given first one (why?), these two are tangent to point of the circle on end points of a diameter (why?).

End the exercise now.

• Thank you ! The line on which the points lie is y=x+sqrt(2) . I also know the tangents are parallel but I could not understand the equation of the line which you have written. Thank you – Seeker1201 May 27 '15 at 13:06
• Could you please explain the equation? – Seeker1201 May 27 '15 at 13:12
• What equation? The line's?: Its slope is $\;-1\;$ and it passes through $\;(0,\sqrt2)\;$,so according to a well knonw formula that is the equation. – Timbuc May 27 '15 at 13:21

Let the given point be $A(0, \sqrt{2})$ which lines on the lies on the line: $y=x+\sqrt{2}$, Now draw the perpendicular say $AN$ from the point $A(0, \sqrt{2})$ to the parallel line: $y=x-2\sqrt{2}$ at the point N.

Thus point N is foot of perpendicular & $AN$ denotes the diameter of circle. Point $N$ is determined by a general expression derived in Reflection Formula by HCR for directly calculating the co-ordinates of foot of perpendicular say $(x', y')$ drawn from any point $(x_o, y_o)$ to the straight line $y=mx+c$ given as $$(x', y')\equiv\left(\frac{x_o+m(y_o-c)}{1+m^2}, \frac{mx_o+m^2y_o+c}{1+m^2}\right)$$ Hence, the co-ordinates of the foot of perpendicular $N$ drawn from the point $A(0, \sqrt{2})$ to the line: $y=x-2\sqrt{2}$ are calculated as follows $$N\equiv\left(\frac{0+1(\sqrt{2}-(-2\sqrt{2}))}{1+1^2}, \frac{1(0)+1^2(\sqrt{2})-2\sqrt{2}}{1+1^2}\right)\equiv\left(\frac{3}{\sqrt{2}}, \frac{-1}{\sqrt{2}} \right)$$ Now, the center of circle is the mid-point of the line AN joining $A(0, \sqrt{2})$ & $N\left(\frac{3}{\sqrt{2}}, \frac{-1}{\sqrt{2}} \right)$ which is given as $\left(\frac{3}{2\sqrt{2}}, \frac{1}{2\sqrt{2}}\right)$ & radius of circle is $\frac{3}{2}$ i.e. half the distance ($=3$) between parallel lines.

Hence, the equation of the circle with center $\left(\frac{3}{2\sqrt{2}}, \frac{1}{2\sqrt{2}}\right)$ & radius $\frac{3}{2}$ $$\left(x-\frac{3}{2\sqrt{2}}\right)^2+\left(y-\frac{1}{2\sqrt{2}}\right)^2=\left(\frac{3}{2}\right)^2$$$$\implies \color{#0b4}{4x^2+4y^2-6x\sqrt{2}-2y\sqrt{2}-4=0}$$

Assume the center to be (a,b). Write the general equation of the circle $(x-a)^2+(y-b)^2=(3/2)^2$ and substitute the point $(0, \sqrt{2})$. This is the first equation.

It is also true that the point (a,b) lies on the the line given by $y = x - \dfrac{\sqrt{2}}{2}$. Substitute for 'a' and 'b'. This is the second equation.

Now you have two unknowns and two equations. Solve it!