Two circles of radius 8 are tangent to the graph of y2 = 4x at the point (1, 2). Find equations of these two circles. (Enter your answers as a comma-separated list.)

Answer 1

Interesting problem. It would look like this:

If both circles are tangent to the function y² = 4x at (1, 2) then they must have the same slope at (1, 2) that y² = 4x has at (1, 2). So step 1 is finding the slope of the tangent line to y² = 4x at (1, 2)

y² = 4x

Taking the derivative implicitly:

d/dx[y²] = d/dx[4x]

2y•dy/dx = 4

dy/dx = 4/(2y) = 2/y

so dy/dx at (1, 2) = 2/2 = 1

Now, lets assume the upper circle, as shown, has a center at (x₁, y₁) and the lower circle has a center at (x₂, y₂). So the equations of the circle are:

(x - x₁)² + (y - y₁)² = 64

(x - x₂)² + (y - y₂)² = 64

If we take the derivative of these, the slope at (1, 2) will be 1 (same as y² = 4x)

so:

d/dx[(x - x₁)² + (y - y₁)²] = d/dx[64]

2(x - x₁) + 2(y - y₁)•dy/dx = 0

dy/dx = -2(x - x₁)/2(y - y₁) = (x - x₁)/(y - y₁)

and at (1, 2), dy/dx = (1 - x₁)/(2 - y₁) = 1 or

(1 - x₁) = (2 - y₁) or

x₁ = y₁ - 1

When we take the derivative of the other circle we get the same thing (except with x₂ and y₂) so:

x₂ = y₂ - 1

Using the equations of the upper circle we developed and substituting in x₁ = y₁ - 1 as well as the point (1, 2) we get:

(1 - (y₁ - 1))² + (2 - y₁)² = 64

(1 - y₁ + 1)² + (2 - y₁)² = 64

(2 - y₁)² + (2 - y₁)² = 64

2(2 - y₁)² = 64

(2 - y₁)² = 32

√(2 - y₁)² = √32

2 - y₁ = ±4√2

y₁ = 2 + 4√2 and y₁ = 2 - 4√2 so:

x₁ = 1 + 4√2 and x₁ = 1 - 4√2 so:

We get the same answers with the other circle. Using our sketch, we can see that the correct combinations of +/- for (x₁, y₁) is (1 - 4√2, 2 + 4√2) and for (x₂, y₂) it is (1 + 4√2, 2 - 4√2)

So the equations of the circles are:

(x - (1 - 4√2))² + (y - (2 + 4√2))² = 64 or

(x - 1 + 4√2)² + (y - 2 - 4√2)² = 64

and

(x - (1 + 4√2))² + (y - (2 - 4√2))² = 64 or

(x - 1 - 4√2)² + (y - 2 + 4√2)² = 64

Answer 2

Hi Celine,

For finding two circles you need to order pair of two center of two circles because radius is 8.

very accurate follow me.

1) graph y2 = 4x and show point (1,2) and two circle through of (1,2) that are tangent to the graph.

2) find tangent line at point (1,2) on the y2 = 4x

so you need derivative ------------------> 2 y y' = 4 then plug in (1,2)

----------------------------------------------------> 2(2) y' = 4 -----> y' = 1 that it means slope of tangent line.

3) write an equation of circle: (x -a )^2 + ( y - b )^2 = r^2 = 64 that center is ( a , b )

4) Find derivative of equation of circle then plug in point (1,2) and y' = m = 1 by doing this you will have an equation between a and b.

(x -a )^2 + ( y - b )^2 = r^2 = 64

2( x - a ) + 2 ( y - b ) y' = 0 ------> 2 ( 1 - a ) + 2 ( 2 - b ) (1) = 0 ----> a + b = 3 #1

5) Now, distance between center ( a , b ) and given point ( 1, 2 ) is radius, so use the formula for finding distance

d = √(a-1)^2 + (b-2)^2 = 8 both side to the power 2

(a-1)^2 + (b-2)^2 = 64 #2

6) Now you need to solve a system include two equations between a and b

a + b = 3 #1 ------> a = 3 - b and plug in to #2

(a-1)^2 + (b-2)^2 = 64 #2 ------------> ( 3 - b -1)^2 + ( (b-2)^2 = 64------> 2 ( b - 2 ) ^2 = 64

b - 2 = ± 4√2 ------> b = 2 ±4√2 then find a

a = 3 - ( 2 ±4√2

Now when b = 2 + 4√2 a = 1 - 4√2 -----> circle: ( x - 1+4√2) ^2 + ( y - 2 - 4√2) ^2 = 64

when b = 2 - 4√2 a = 1 +4√2 -------> circle: ( x - 1-4√2) ^2 + ( y - 2 +4√2) ^2 = 64

I hope it is useful.

Minoo

Answer 3

The slope of the tangent line is 1; therefore, the slope of the perpendicular is -1.

The sides of the right triangle with hypotenuse = segment between the center and the point of tangency form an isosceles right triangle.

The sides of a isosceles right triangle with hypotenuse of 8 are both 4sqrt(2).

One center is [1-4sqrt(2),2+4sqrt(2)].

The other center is [1+4sqrt(2),2-4sqrt(2)]