Question

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?

Answer 1

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.

Answer 2

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

Answer 3

nice one..campbell_st

Answer 4

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

Answer 5

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

Answer 6

Fair enough.