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Q: What is Circle minus square equals triangle?

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If ... the square of (the x-coordinate of the point minus the x-coordinate of the center of the circle) added to the square of (the y-coordinate of the point minus the y-coordinate of the center of the circle) is equal to the square of the circle's radius, then the point is on the circle.

Plus or minus the base. If the base is X and you square it, you get X2. If you take the square root of that, you get Plus or Minus X. This is because X*X equals X2 and -X*-X also equals X2.

A equals Vf minus Vi divided by time equals triangle v divided by time

No. But sin2a equals 1 minus cos2a ... and ... cos2a equals 1 minus sin2a

square root of 8

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180

If ... the square of (the x-coordinate of the point minus the x-coordinate of the center of the circle) added to the square of (the y-coordinate of the point minus the y-coordinate of the center of the circle) is equal to the square of the circle's radius, then the point is on the circle.

Plus or minus the base. If the base is X and you square it, you get X2. If you take the square root of that, you get Plus or Minus X. This is because X*X equals X2 and -X*-X also equals X2.

A equals Vf minus Vi divided by time equals triangle v divided by time

No. But sin2a equals 1 minus cos2a ... and ... cos2a equals 1 minus sin2a

If x equals the square root of ...., then you already have solved for x

square root -5 minus 14 or - square root -5 minus 14

Equation of circle: x^2 +y^2 -10y -24 = 0 Completing the square: x^2 +(y-5)^2 = 49 Center of circle: (0, 5) Radius of circle: 7 Distance from (7, -2) to (0, 5) is 7*square root of 2 is hypotenuse of right triangle Using Pythagoras theorem: hypotenuse squared minus radius square = 49 Length of tangent line is the square root of 49 which is 7

Yes it does. Good work!

square root of 8

5 times 2 square equals 20 minus 29 equals -9

The area enclosed by a chord is equal to the area enclosed by a segment minus the area enclosed by the triangle with the same corners as the segment. To visualise it, draw a circle and put a chord on it. Label the chord AB and the centre of the circle C. The area of sector AB equal to the area of sector ABC minus the area of triangle ABC.