My Math Forum Circle problem

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 March 12th, 2014, 03:49 AM #1 Newbie   Joined: Jul 2013 Posts: 21 Thanks: 0 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? Please help. Thank you
 March 12th, 2014, 05:11 AM #2 Member   Joined: Mar 2013 Posts: 90 Thanks: 0 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 $a$.
 March 12th, 2014, 08:16 AM #3 Newbie   Joined: Jul 2013 Posts: 21 Thanks: 0 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..
 March 12th, 2014, 09:22 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,307 Thanks: 1976 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.
 March 12th, 2014, 10:27 AM #5 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,600 Thanks: 2588 Math Focus: Mainly analysis and algebra 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.
 March 12th, 2014, 10:39 AM #6 Global Moderator   Joined: Dec 2006 Posts: 20,307 Thanks: 1976 That's why my answer works, as symmetry implies those distances are equal. As the lines are not parallel, their normals must intersect.
 March 12th, 2014, 12:59 PM #7 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,600 Thanks: 2588 Math Focus: Mainly analysis and algebra 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!
 March 19th, 2014, 11:57 PM #8 Senior Member   Joined: Mar 2014 Posts: 112 Thanks: 8 I got (x - 4)² + (y - 4)² = 10.
 March 20th, 2014, 01:02 AM #9 Senior Member   Joined: Mar 2014 Posts: 112 Thanks: 8 Also, if the circle doesn't require having such tangent lines then (x + 2)² + (y - 4)² = 10.
 March 21st, 2014, 07:04 PM #10 Senior Member   Joined: Mar 2014 Posts: 112 Thanks: 8 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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# show that a circle can be drawn touching the line

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