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Consider two lines having equations : $u:2x+y-5=0 \\ v:x+2y+2=0$ Then , find the equation(s) of the circle which touches each of the line $u,v$ and has radius $\sqrt{5}$ .

Nice hints and solutions are always welcome!

#Geometry #Help

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Comments

Idea: Find the lines parallel to the given lines at a distance of $\sqrt5$ , and intersect them. Those lines are $2x+y-5=\pm5$ for u and $x+2y+2=\pm5$ for v. Just solve those four simple linear equations to find the centers of the four circles..

Otto Bretscher - 6 years, 1 month ago

Since radius of the circle is given we need to find the center. Let the center be $\left( h,k \right)$ ,then $\displaystyle \cfrac { \left| 2h+k-5 \right| }{ \sqrt { 5 } } =\cfrac { \left| 2k+h+2 \right| }{ \sqrt { 5 } } =\sqrt { 5 }$ . Find $\left( h,k \right)$ then apply standard formula to write the equation of the the circle.

Soumo Mukherjee - 6 years, 1 month ago

I understood it. Indeed, equating the distances only gives the equation of angle bisectors.@Soumo Mukherjee @Otto Bretscher Thanks for you help and time.

Nihar Mahajan - 6 years, 1 month ago

You are using equation of angle bisectors right?

Nihar Mahajan - 6 years, 1 month ago

Distance formula. I don't remember the formula for angle bisector. However, it is the locus of the point equidistant from both the arms of the angle. So you can interpret it that way, too.

@Otto Bretscher suggestion is pretty cool and clever : there's always a better way.

Soumo Mukherjee - 6 years, 1 month ago

@Soumo Mukherjee – Our solutions are equivalent...

Otto Bretscher - 6 years, 1 month ago

@Otto Bretscher – However the way you interpreted was really nice: Finding a point as a intersection of two lines is the simplest in coordinate geometry. And the center is a point.

I will use this interpretation in other future problems

Soumo Mukherjee - 6 years, 1 month ago

Sir Otto Bretscher has already said what I wanted to say. Anyway the coordinates of center of four circles are $(\dfrac{17}{3},\dfrac{-4}{3}),~(9,-8),~(-1,2)~ \& ~(\dfrac{7}{3},\dfrac{-14}{3})$ .

But I can help you finding the angle bisector of two lines (or bisector of two planes in case of 3D Coordinate Geometry).

The equation of angle bisector of lines $a_1x+b_1y+c_1=0 ~\& ~ a_2x+b_2y+c_2=0$ can be written as $\\ \dfrac{|a_1x+b_1y+c_1|}{a_1^2+b_1^2} = \dfrac{|a_2x+b_2y+c_2|}{a_2^2+b_2^2} \\ \text{or} \dfrac{a_1x+b_1y+c_1}{a_1^2+b_1^2} = \pm \dfrac{a_2x+b_2y+c_2}{a_2^2+b_2^2}$ .

But you don't have to remember this formula as you can derive it anytime by finding the locus of a point which is equidistant from both lines.

Purushottam Abhisheikh - 6 years, 1 month ago

Thanks for your help and time.

Nihar Mahajan - 6 years, 1 month ago

@Calvin Lin @Brian Charlesworth @Jon Haussmann @Tanishq Varshney @Purushottam Abhisheikh @Sharky Kesa @Trevor Arashiro @Krishna Sharma @Curtis Clement @Pi Han Goh @Chew-Seong Cheong @everyone please help.

Nihar Mahajan - 6 years, 1 month ago

Please reserve @ mentions for targeting of specific people when you know that they will be interested. For notes like this, just let it appear naturally in their feed, they will reply if they see it. Avoid mass targeting @ mentions like this, and do your best to limit it to under 5 people.

Calvin Lin Staff - 6 years, 1 month ago

Actually I had seen many people when needed help , tagging many people in their note.If it is causing trouble for you and some people , I will stop tagging many and limit my tags to 5 people. Thanks!

Nihar Mahajan - 6 years, 1 month ago

@Nihar Mahajan – Right. I've began to receive complaints about the abuse of this functionality, and I agree that it can be extremely irritating for all of those involved. (EG For the 11 people that you tagged to land on this page, only to realize that 1) they are not interested in this, or 2) others have already responded.) I will be responding in a similar manner to such instances in future. I would prefer for the community to understand what is proper etiquette, rather than trying to impose rules on what can or cannot be done in the product.

It is alright if you tag a specific person and say something like "Hey, you helped me out with a similar problem in the past. I did the following work and then got stuck. Any suggestions for what else to do?". In this way, the person that is tagged will understand the context, and be more willing to assist you.

Calvin Lin Staff - 6 years, 1 month ago

Let me post my approach :

First we must notice that there exist such $4$ circles. There are $2$ angle bisectors of the angle between these two lines which are the locus of the $4$ centers of these circles.For this we need to know the equations of angle bisectors.So , here is the problem for me. I am finding it difficult to write the equations of angle bisectors.Once the equation of angle bisectors are known , problem is solved.

Nihar Mahajan - 6 years, 1 month ago

See the aim is to find the equation of the circle. Equation of the circle can be found if we know its center and its radius. We already have the radius. So you need only the center. So next things that pops is that can the center be found easily with what is given?

The center lies of the angle bisector: acute or obtuse (you need to confirm this).

The equation of the angle bisector can be written if you know the distance formula: distance of a point from a line. The angle bisector is the locus of the point equidistant from both the arms of the angle. Or you can look it up (the formula) elsewhere. But do you really need the formula or the acute angle bisector formula? or that of the obtuse angle.

The circle is uniquely defined since it is satisfying three given conditions. You need to search only one such circle, not 4 such circles.

Soumo Mukherjee - 6 years, 1 month ago

Generally I find it helps to accurately draw the graphs to at least get an idea of what answer you will get.

Curtis Clement - 6 years, 1 month ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$