Find the equation of the circle which touches the line $3x+y+3=0$ at $(-3,6)$ and tangent to the line $x+3y-7=0$
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1Welcome to math.SE, since you appear to be new I wanted to give you a couple of tips. First of, it is usually helpful to say in what context this problem was found, and more importantly which are your thoughts. Second, some of us consider imperative expressions such as "find", "prove'', "evaluate'' ... somewhat rude, try to be polite. And finally, perhaps related with the first point, this is not a site for solving homework, please consider this and try to rephrase your question – caverac Sep 22 '17 at 12:42
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@Harshit Pant Something is wrong in your given. – Michael Rozenberg Sep 22 '17 at 12:56
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Nope it is as it is. I am posting the answer graph in the question too now. – Harshit Pant Sep 22 '17 at 13:03
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You've got $3$, where you mean $-3$. – Jam Sep 22 '17 at 13:10
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You seem to be missing some information. There's no unique solution to this problem, in fact you get a whole family of circles. Presumably, you also have the criterion that the circle is tangent to $3x+y+3=0$. The following hint will cover the tangent case and the general case. I'm not convinced that you'll read all this but I'll leave it up for posterity and my own practice.
Hint 1: For the case that the circle is tangent to the line $3x+y+3=0$ at $(-3,6)$. Take the general equation of a circle centered at $(a,b)$, which is $(x-a)^2+(y-b)^2=r^2$. You can find a perpendicular line to $x+3y-7=0$ by noting that the product of the gradients $m_1m_2=-1$. Do the same for the line $3x+y+3=0$. Find their point of intersection, as $(a,b)$ and deduce the formula of the circle using the knowledge that $(-3,6)$ satisfies the equation.
Hint 2: For the general case, where the circle touches the point but isn't tangent. The line, $l$, which is perpendicular to $x+3y-7=0$ will be $y=3x+c$ for some constant, $c$. The lower point where $l$ meets the circle is then $\left(\frac{7-c}{10},\frac{21+c}{10}\right)$. The circle touches both this point and $(3,-6)$ so the Euclidean distance from $(a,b)$ to $(-3,6)$ is equal to the Euclidean distance from $(a,b)$ to $\left(\frac{7-c}{10},\frac{21+c}{10}\right)$ as they're both radii of the circle. The line $l$ is tangent to the circle, so it passes through the circle's center, so $b=3a+c$. Using these facts and setting these distances equal to each other, you can find that $a$ is equal to a quadratic in $c$. We can then find $r$, by the formula for Euclidean distance and find $b$ with the line $l$.
Thus, the circle becomes dependent on the constant $c$. In fact, when $c=0$, the problem degenerates to the simpler case in hint 1. The figure below shows an example of one such circle. Note that the circle intersects the line $3x+y+3=0$ twice, since it's not tangent. If you go through the algebra of this, be careful not to drop a minus; I did so and was confused for a long time.
Jam
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