0
$\begingroup$

The circles $C_1$ and $C_2$ are given by
$x^2+(y+3)^2=1$
and
$x^2+y^2-12x-10y-60=0$

C1 has a centre of (0,-3) and radius of Square root 1, whereas C2 has a centre of (6,5) and radius of 11!

  • 0
    What have you tried? Have you found the centers of $C_1$ and $C_2$? That is an important starting point. – 2017-02-12
  • 1
    Yes I have found that the centre and radius of C1 is (0,-3) and square root 1 for the radius. – 2017-02-12
  • 0
    I have also found that the centre for C2 is (6,5) and its radius is 11. I dont know what steps to carry through after though! – 2017-02-12

2 Answers 2

2

As you mentioned in the comments, you'd first find the centers of these circles.

So the center of $C_1$ is $(0,-3)$ and the center of $C_2$ is $(6,5)$.

Start by drawing these points on some graph paper and sketching circles that go through both points. What happens to the center of your new circle as it gets smaller?

You should note that the center of every circle that goes through both of these points is on the perpendicular bisector of $(0,-3)$ and $(6,5)$. That's because if both points are on the circle, we can draw a chord between them and the perpendicular bisector of that chord will go through the center.

Hopefully you'll conclude that the smallest circle going through both points is centered at $(3, 1)$. Can you see what's special about that point?

  • 0
    Ok thank you very much! I will plot the circles on the graph now – 2017-02-12
  • 0
    Hello again, I have plotted the 2 circles on a graph and now can see the gap between them, but I am now confused on how I will be able to work out the coords and radius for it as the circles are not overlapping and using measurements from the graph would not be valid! – 2017-02-12
  • 0
    From your question, you're just looking for a circle that passes through the centers of $C_1$ and $C_2$, right? so it doesn't actually matter what $C_1$ and $C_2$ look like, just where there centers are. You want to plot just the centers, and draw circles that DO go through both of them. – 2017-02-12
  • 0
    OHH ok right thank you so much, I will try that now x – 2017-02-12
  • 0
    This might help: https://www.desmos.com/calculator/wsnljozo9v You can see the two points (0,-3) and (6,5), and then drag the black point to change the center of the circle that goes through those two points. Do you see when it's smallest? – 2017-02-12
  • 0
    Right, I have just done that now and it gave the results with coords of (3,1) with a radius of 10cm. Would you know just by the sound, that that is right or not? – 2017-02-12
  • 0
    Almost - I think you perhaps mean the diameter is 10? – 2017-02-12
  • 0
    No, my graph is showing that is has a radius of 10 and a diameter of 20, so i dont know what has gone on! – 2017-02-12
  • 0
    If the center is (3,1) and (0,-3) is a point on the circle, then what's the radius? $r = \sqrt{(3-0)^2 + (1-(-3))^2} = \sqrt{3^2 + 4^2} = 5$. – 2017-02-12
  • 0
    Well, I would agree with that equation and therefore will, but my graph is suggesting that it is 10. Anyways, thank you so much!! – 2017-02-12
0

The smallest circle will lie on the midpoint of the line connecting the centres, and will have diameter of the length of that line. Since you know how to construct and interpret the circle equations, you should be able to complete the exercise with that knowledge.

In response to comment request: For a circle centre $(a,b)$ and square of radius $r^2$, the equation is $(x-a)^2 + (y-b)^2=r^2$. The square of the radius is often (as here) what is available from review of coordinates; there may be no need to calculate down to the actual radius.

  • 0
    I suppose but I really am confused on just how I work out the equation for the circle, could you demonstrate it for me please x – 2017-02-12
  • 0
    @JosefA updated – 2017-02-12