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 shiseonji March 12th, 2014 02:49 AM

Circle problem

Show that a circle can be drawn touching the line 3x+y-6=0 at (1, 3) and 3x-y+2=0 (1, 5).

So tell me if I'm wrong but what I'm supposed to do is to find the distance of both lines from the unknown point of the circle by using the directed distance formula from a point to a line, Ax+Bx+C / sqrt(A^2+B^2) to both the equations and equate the two expressions. So if I do that, I will have this:

$\left|\frac{3x+y-6}{\sqrt{10}} \right|=\left |\frac{3x-y+2}{-\sqrt{10}} \right |\$

And if I solve for h, which is equal to $\frac{2}{3}$, I won't get k.

So, how do I do this right?

 Nehushtan March 12th, 2014 04:11 AM

Re: Circle problem

The centre of the circle is clearly some point $(a,\,4)$. The line joining this and the point $(1,\,3)$ is orthogonal to the line $3x+y-6=0$. A vector parallel to that line is $\begin{pmatrix}-1 \\ 3\end{pmatrix}$. Thus
• $\begin{pmatrix}-1 \\ {} \\ 3\end{pmatrix}\cdot\begin{pmatrix}a-1 \\ {} \\ 4-3\end{pmatrix}\=\ 0$

and you can solve for $a$.

 shiseonji March 12th, 2014 07:16 AM

Re: Circle problem

Hello. Thank you for responding. But can you explain it in a way that it doesn't involve vectors? We haven't really studied about them yet.. and how did you know it is some point (a, 4)? and what does $\begin{pmatrix} -1\\ 3\end{pmatrix}\$ mean? Does that mean the point (-1, 3)? Sorry.. I'm so poor at this stuff.. :(

 skipjack March 12th, 2014 08:22 AM

The given lines intersect at the point (2/3, 4). As their gradients are -3 and 3, the lines are symmetrical about lines parallel to the axes through (2/3, 4). The relevant one of those lines is the line y = 4, because the points (1, 3) and (1, 5) are symmetrical placed in relation to the line y = 4. The symmetry implies the circle exists and that lines drawn through the points (1, 3) and (1, 5) perpendicular to the lines 3x + y - 6 = 0 and 3x - y + 2 = 0 will intersect at a point on the line y = 4 which is the centre of the circle. You don't need to calculate that point's coordinates.

 v8archie March 12th, 2014 09:27 AM

Re: Circle problem

I would observe that, if the two lines touch the circle at the given points, this means that they are tangents to the circle at those points.Thus, the normals to the circle are the perpendiculars to the two lines at the points given. The centre of the circle must then be where those normals intersect.

Thus, if you show that the normals intersect and that the point of intersection is equidistant from the two points given (the radius of the circle) then you are done.

 skipjack March 12th, 2014 09:39 AM

That's why my answer works, as symmetry implies those distances are equal. As the lines are not parallel, their normals must intersect.

 v8archie March 12th, 2014 11:59 AM

Re: Circle problem

Ah, yes. When first I browsed your solution it wasn't immediately clear to me what you were driving at. It all makes perfect sense in the clear light of day!

 the john March 19th, 2014 10:57 PM

I got (x - 4)² + (y - 4)² = 10.

 the john March 20th, 2014 12:02 AM

Also, if the circle doesn't require having such tangent lines then (x + 2)² + (y - 4)² = 10.

 the john March 21st, 2014 06:04 PM

Notice that its centers are (4, 4) and (-2, 4) respectively with radius ?10.
Also x² + y² - 8x - 8y + 22 = 0 and x² + y² + 4x - 8y + 10 = 0 respectively.

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