Circle Equation

Question

Find an equation of the circle with center at (4,-6) that is tangent to the y-axis in the form of (x-A)^2+(y-B)^2=C where A, B, C are constant. 

Answer 1

Use (x-h)² + (y-k)² = r² for the equation of a circle with center (h,k) and radius r.

For the circle to be tangent to the y-axis  with a center at (4,-6) requires the radius to be 4.

The equation would be (x-4)² + (y+6)² = 4²

To match the format you specified, replace 4² with 16.

 (x-4)² + (y-(-6))² = 16

Answer 2

The standard equation for a circle is (x-A)²+(y-B)²=C

 

where the center is the point (A,B) and C is the radius squared.

 

you have been given the center so that is just a matter of plugging in the values

 

(x-(4))²+(y-(-6))²=C

 

To find the radius it will help to draw a sketch. It will help if the circle in your sketch is as round as possible so use a compass if possible or trace a quarter or a bottle cap.

 

 

Remember tangent means it touches the axis at exactly one point and the radius is the distance from the center to any point on the circle. (the one point  of your circle on the y axis is a good one to use to find the radius)

 

Comment if you need more help.

Answer 3

Sally,

 

Point (A, B) is just the center of the circle given as (4, -6), and C is the radius squared.

Given that the circle is tangent to the y-axis (meaning x=0} and that a circle's tangent is always perpendicular to its radius, it must just touch the y-axis at point (0, -6).

 

So the equation becomes (x-4)² + (y+6)² = (4)² = 16   (i.e, A=+4, B=-6 & C=16)

 

To check its tangency at any point (x,y) on the circle (if you know how to take derivatives), compute dy/dx by taking the derivative of both sides of the equation and solving for dy/dx:

 

2(x-4)dx/dx + 2(y+6)dy/dx = 0,  So  dy/dx = -2(x-4)/2(y+6) = (4-x)/(y+6) = tangent expression.

 

So at point (0, -6) on the y-axis   dy/dx = 4/0 or infinity; hence, its slope or tangent is the y-axis.  It checks out!

At the opposite point (8, -6) on the circle, dy/dx = -4/0 = - infinity, so its tangent is parallel to the minus y-axis.

At points (4, -2) and (4, -10), dy/dx = 0, so the tangents are parallel to the x-axis as lines y=-2 and y=-10.