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Let us suppose there are two straight lines having equation x=5 and y=7 and a circle is drawn such that these two straight lines are tangents to the circle. Now we are required to find the center of the circle. How to find it? I mean aren't there four different circles possible? Which circle's center is the question asking about?

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    $\begingroup$ Not only that, there's also the question of how large the circle is... $\endgroup$ – angryavian Jan 8 '14 at 15:46
  • $\begingroup$ If a circle is tangent to two straight lines, this means that its center belongs to one of the bisectors of the four angles formed between these two lines, and its radius is obviously the distance from this point to each of the two lines. $\endgroup$ – Lucian Jan 9 '14 at 3:36
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It is not possible to find the center of the circle, because (as mentioned in the comments before) there is an infinite amount of possible circles. But all of them have some property in common:

  • The center lies either on the line defined by $\left\lbrace (x,y) : (5,7)+r\cdot(1,1), r\in\mathbb{R}\right\rbrace$ or
  • The center lies on the line defined by $\left\lbrace (x,y) : (5,7)+r\cdot(1,-1), r\in\mathbb{R}\right\rbrace$

If $p$ is point on one of those lines, then radius of this circle with center $p$ is given by

$$\frac{1}{\sqrt{2}}\left\|p-(5,7)\right\|$$

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Suppose we have two straight lines, $x=5$, and $y=7$. We want the center of the circle of radius $r$ so that both lines are tangent to the circle. While there are infinitely many such circles with arbitrary radius $r$, we can find them by noting that the center of the circle will be a horizontal distance of $5 \pm r$ from the line $x=5$, and it will also be a vertical distance of $7 \pm r$ away from the line $y=7$. This leaves 4 possible equations for this circle. Namely,

$$\left(x-\left(5+r\right)\right)^2+\left(y-\left(7+r\right)\right)^2=r^2,$$

$$\left(x-\left(5-r\right)\right)^2+\left(y-\left(7+r\right)\right)^2=r^2,$$

$$\left(x-\left(5-r\right)\right)^2+\left(y-\left(7-r\right)\right)^2=r^2,$$

and

$$\left(x-\left(5+r\right)\right)^2+\left(y-\left(7-r\right)\right)^2=r^2$$

The centers are respectively, $$(5+r,7+r),(5-r,7+r),(5-r,7-r), \operatorname{and} \,(5+r,7-r).$$

I took the liberty of making a dynamic slider graph of this effect for you. You can change the radius $r$ around and see how the 4 circles meet your request. Leave $x=5$ and $y=7$ alone in this graph. If you change them you need to change the equations accordingly, I did not account for them changing in the circle equations (but we could ;)).

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  • $\begingroup$ Thanks for your remark. :+) $\endgroup$ – mrs Jan 8 '14 at 17:38
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As with any two non-coincident straight lines you have infinite possible circles. The question may be referring to the geometrical place of those circles, or maybe you are missing something in your question, like the radius of the said circle...

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  • $\begingroup$ No, the question gives only this much information and nothing more than that and it only asks for the center, this is an extract from SAT subject test Mathematics! $\endgroup$ – Avery Jan 8 '14 at 15:49
  • $\begingroup$ @user32340: U mean the problem just asks about the centre? $\endgroup$ – mrs Jan 8 '14 at 15:52
  • $\begingroup$ Well, then your question is either incomplete or the examiner is being intentionally broad to see how you will handle the problem. I'm not American, so let me ask: Aren't SAT multiple choice tests? If so, maybe you are supposed to see which of the option represents a possible center of a tangent circle, not "the" center, as we already now that there are infinite of those... $\endgroup$ – Zado Jan 8 '14 at 15:58
  • $\begingroup$ A SAT question asking a nonsense...hmmm. $\endgroup$ – DonAntonio Jan 8 '14 at 16:03
  • $\begingroup$ @Zado Yes I too had a very tough time solving this before having adopted hit and trial method from the choices given!! $\endgroup$ – Avery Jan 8 '14 at 16:08

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