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One of my math tests has this question.

A circle has its center at $(6,7)$ and goes through $(1,4)$. Another circle is tangent at $(1,4)$ and has the same area.

What are the possible coordinates of the second circle. Show your work or explain how you found you answer.

Could someone solve a sample question to show me how to solve this one?

Community
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Joe
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  • Find the line through (6,7) and (1,4). The centre of your second circle is on this line, the same distance as from (6,7) to (1, 4) but on the other side of the tangent line. Find that distance (a radius) and that centre. – Paul Mar 05 '15 at 16:20
  • Hint: The center of the second circle has to lie on the line defined by the points $(6,7)$ and $(1,4)$ (otherwise the circles are not tangent), and it has to have the same distance from $(1,4)$ as $(6,7)$ (otherwise the circles do not have the same radius, and therefore area).

    Edit: sniped :|

     –  Mar 05 '15 at 16:20
  • Sorry to say that this question is badly phrased for the following reasons. It lacks accurate details like those included in the brackets below. 1. Another circle is tangent (to the given circle) at (1,4). 2. What are the possible coordinates of (the center of) the second circle. 3. By “*coordinates*”, it refers to x and y coordinates of the center of the second circle, and that is why it is in plural. It tries to mislead you to think that there are more-than-one circles. Judging from the title of your post, it seems you have been ….. – Mick Mar 05 '15 at 17:04

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The center $Q=(h,k)$ of the second circle lies on the line defined by $P_1=(6,7)$ and $P_2=(1,4)$. What means

$$\frac{k-4}{h-1}=\frac{4-7}{1-6}=\frac{3}{5}...(1)$$

Since these circles have the same area it follows they have equal radius. Then $$\sqrt{(h-1)^2+(k-4)^2}=\sqrt{(1-6)^2+(4-7)^2}...(2)$$

Solving the system of equations formed by $(1)$ and $(2)$ you find the center of the second circle.

Ángel Mario Gallegos
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