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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?

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  1. anonymous
    • 4 years ago
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    |dw:1328695605618:dw|

  2. anonymous
    • 4 years ago
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    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
    • 4 years ago
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    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
    • 4 years ago
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    nice one..campbell_st

  5. anonymous
    • 4 years ago
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    @campbell_st : how are you so sure about the fact that the circle lies in the first quadrant?

  6. campbell_st
    • 4 years ago
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    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
    • 4 years ago
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    Fair enough.

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