## anonymous 4 years ago What is the equation of the circle if its radius is 5 and its center is 12 units away from the origin bearing 45 degrees?

1. anonymous

|dw:1328695605618:dw|

2. anonymous

sin(45) = AB / OB 0.707 = AB / 12 AB = 12*0.707 AB = 8,49 => Yb = 8.49 cos(45) = OA / OB 0.707 = OA / 12 OA = 8.49 => Xb = 8.49 etc.

3. campbell_st

so the hypotenuse is 12 units... the triangle formed by ABO is a right isosceles... with 45 degree angles... using pythagoras' thorem a^2 + a^2 = 12^2 2a^2 = 144 so $a= 6\sqrt{2}$ so the centre is $(6\sqrt{2}, 6\sqrt{2})$ using the general form of the circle $(x - h)^2 + (y-k)^2 = r^2$ the circle is $(x - 6\sqrt{2})^2 + (y - 6\sqrt{2})^2 = 5^2$ you can expand and simplify it

4. anonymous

nice one..campbell_st

5. anonymous

@campbell_st : how are you so sure about the fact that the circle lies in the first quadrant?

6. campbell_st

its an angle bearing of 45... Bearing are measured from Nth in a clockwise direction. The diagram above should have the angle from the y-axis.... and not the x-axis... but thats not a big issue

7. anonymous

Fair enough.