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Circle $C_1$ has the equation $x^2 + y^2 - 18x - 22y -23 = 0$ and circle $C_2$ has the equation $x^2 + y^2 + 14x + 2y + p = 0$

Find the value of $p$ if the circles touch at one point and find the point of contact.

I have solved that p = 25; I just need point of contact.

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    What have you attempted **yourself** ? – 2017-02-27
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    i know p = 25 that is all – 2017-02-27
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    Questions posted to math.stackexchange.com should not be phrased in language suitable for assigning homework. That can make people wonder if someone copied a question without understanding it. You can ask for help understanding a question, or you can say specifically what difficulties you had solving a problem, but merely copying is frowned on. – 2017-02-27
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    Then, fraser, you should add that $p = 25$ *into your question post directly*. Note that you ask "Find the value of p...." – 2017-02-27
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    I am using the question as an example as i do not know the technique for solving this style of question. I am just trying to learn – 2017-02-27

1 Answers 1

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If we convert the equations of the circles to standard form, the first equation becomes $$(x-9)^2-81 + (y-11)^2-121 - 23 = 0$$ $$(x-9)^2 + (y-11)^2 = 15^2,$$ and the second equation reduces to $$(x+7)^2-49 + (y+1)^2-1 + p = 0$$ $$(x+7)^2 + (y+1)^2= (\sqrt{50-p})^2.$$ The distance between the two centers is $\sqrt{(9-(-7))^2+(11-(-1))^2}=20$. If the two circles are tangent, this means the distance between the centers is equal to the sum of the two radii; in other words, $20=15+\sqrt{50-p}$. Solving for $p$, we get $p=25$.

To find the point of tangency, The key realization is that the point of tangency lies on the line segment connecting the center of the two circles, $(9, 11)$ and $(-7, -1)$. Note this line segment has length $20$. Because the radius of $C_1$ is $15$, the point of tangency must lie $\frac{15}{20}$ of the way from $(9, 11)$ to $(-7, -1)$. Therefore the tangent point is $(-3, 2)$.

Alternatively, the two equations $(x-9)^2 + (y-11)^2 = 15^2$ and $(x+7)^2 + (y+1)^2= 5^2$ could have been solved as a system of simultaneous equations, yielding the same answer.

Graph of the two equations, three points, and line segment

Credit to Desmos for the graph

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    yeah. i managed to get that however the point of contact i could not find. – 2017-02-27
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    thank you appreciate the help. Didnt realise you could solve through simultaneous equations – 2017-02-27
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    No problem! I also added an image. – 2017-02-27