Equations of circles tangent to graph

In summary: Then use b= 3- a to find b.In summary, the two circles with radius 3√2 are tangent to the graph y^2 =4x at (1,2). To find the equation of each circle, the slope of the tangent line at (1,2) must be found by taking the derivative. The slope of the perpendicular line through (1,2) can then be determined, which is also the slope of the circles. Using the distance formula, the centers of the circles can be found to be a distance of 3√2 from (1,2). Finally, by solving for the equations of the circles, the final solution can be obtained.
  • #1
emma3001
42
0
Two circles of radius 3√2 are tangent to the the graph y^2 =4x at (1,2). Find the equation of each circle.

I have found the derivative of the graph, which is 1/√x. I know that the equation of the circle is X^2 + Y^2=r^2 where r is the radius so the equation of each circle is X^2 + Y^2=18. That is all I can get. Please help!
 
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  • #2
emma3001 said:
Two circles of radius 3√2 are tangent to the the graph y^2 =4x at (1,2). Find the equation of each circle.

I have found the derivative of the graph, which is 1/√x.
Okay, what is that at (1, 2)?

I know that the equation of the circle is X^2 + Y^2=r^2 where r is the radius so the equation of each circle is X^2 + Y^2=18.
No, you don't know that. That is the equation of a circle with center at (0,0) and one thing you can be pretty sure of is that these circles do NOT have center at (0,0)!

That is all I can get. Please help!
If you have found the derivative at (1, 2) then you know the slope of the tangent line. What is the slope of, and what is the equation of, the line through (1, 2) perpendicular to the tangent line? I ask that because a tangent to a circle is always perpendicular to a radius. The centers of your two tangent circles must lie on the line through (1, 2) perpendicular to the tangent line to the curve, at distance [itex]3\sqrt{2}[/itex].
 
  • #3
okay so i have made progress...

I plugged in x=1 into the derivative so the slope of my tangent is 1. Then I found the equation of the tangent which is y=x + 1. I now know that the slope of my circles is -1 and I know that
(x-a)^2 + (y-b)^2= 18

Could you help me with the next step? In other words, how do I find a point with that slope and a distance or radius of 3sqrt2?
 
  • #4
emma3001 said:
okay so i have made progress...

I plugged in x=1 into the derivative so the slope of my tangent is 1. Then I found the equation of the tangent which is y=x + 1. I now know that the slope of my circles is -1 and I know that
(x-a)^2 + (y-b)^2= 18

Could you help me with the next step? In other words, how do I find a point with that slope and a distance or radius of 3sqrt2?
The slope of the circles is not -1; this is the slope of the line perpendicular to the tangent.

Once you've found the equation of this perpendicular line...

The cool thing is that the slope of the perpendicular line is -1 so the change in the x and change in the y are the same. Hint: If you draw a triangle you will see that it is a special triangle. From here, solve for the side(s) of that triangle.

Once you've done that, just simply add/subtract to (1,2) and you will get the centers of both circles. Finally, solve for the equation of the circles.
 
  • #5
i know that my equation of the line perpendicular to the tangent is y=3-x. From that, do I use my distance formula to get the centre of the circle... before it said something about special triangles- could you elaborate?
 
  • #6
The centers of your two circles, (a, b), are ON the line y= 3- x, so you know b= 3- a. Further, they are distance [itex]3\sqrt{2}[/itex] from (1, 2). The distance from (a,b) to (1, 2) is [itex]\sqrt{(a-1)^2+ (b-2)^2}= \sqrt{(a-1)^2+ (3-a-2)^2}[/itex]
Set that equal to [itex]3\sqrt{2}[/itex] and solve for a.
 

1. What is the equation of a circle tangent to a graph?

The equation of a circle tangent to a graph is given by (x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center of the circle and r is the radius.

2. How do you find the center of a circle tangent to a graph?

The center of a circle tangent to a graph can be found by finding the point of tangency between the circle and the graph. This point will be the center of the circle.

3. Can a circle be tangent to a graph at more than one point?

No, a circle can only be tangent to a graph at one point. This is because a circle has only one radius, and if it were to be tangent at more than one point, the radius would have to be different at each point.

4. How do you determine the radius of a circle tangent to a graph?

The radius of a circle tangent to a graph can be found by using the distance formula to find the distance between the center of the circle and the point of tangency. This distance will be the radius of the circle.

5. Can the equation of a circle tangent to a graph have a negative radius?

No, the radius of a circle must always be a positive number. If the equation of a circle tangent to a graph results in a negative radius, it means that the circle does not actually intersect the graph at any point, and therefore is not tangent to it.

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