A circle of radius 5 is centered at the point (-6,7).?

Question

Find a formula for f(theta), the y-coordinate of the point P=(x,y) on the circle specified by the angle (theta). Think about trig functions. I can't draw the circle that goes with this problem but the hyp looks to make a 45degree angle. I just don't understand how to go about doing this. Any help would be great!

I need to find a formula for f(theta), which is the y-coordinate of the point P=(x,y).

Answer 1

This type of problem will eventually seem EASY for you, so hang in there. Let's begin by stressing that "a circle with radius 5" is simple to draw -- I recommend marking the 4 points that are "5 units" above, below, right and left of the center point.

That is, mark (-6,7) on your graph paper, then sketch four more points at (-1,7), (-11,7), (-6, 2) and (-6,12). "Connect the dots" by drawing all 4 quarter-circle arcs between the points.

The equation for any circle -- in "rectangular coordinates" (which means (x,y)) -- is normally: (x-a)^2 + (y-b)^2 = r^2, where (a,b) is the center point and r is the radius.

But what you're hoping to find is the equation of a circle in "polar coordinates" (which means (r,theta)).

The reason a sample triangle was drawn inside the given circle (I'm assuming you're looking at a textbook) is to illustrate the relationships among x, y, theta and r. Let's inspect, using rules from trignometry:

x = r cos (theta) + a
y = r sin (theta) + b
r = sqrt ((x-a)^2 + (y-b)^2)
theta = Arctan ((y-b)/(x-a))

where (x,y) is any point on the circle, (a,b) is the circle's center, and r is the radius.

The formula for theta that I just provided may be precisely what you are looking for. It works simply by first finding the "slope" of the line connection (x,y) and (a,b), then taking the inverse tangent of that slope, which 'converts' it to an "angle."

Hope this helps!

Answer 2

For the first part
Eq of the circle
(x-(-6))^2+(y-7)^2=5^2
(x+6)^2+(y-7)^2=25
solve this
for the second part
specify if it is a continuation of first part