Find the point on the curve x² + y² = 144 closest to (0, 16). A) (0, 144) B) (0, 12) C) (0, 16) D) (12, 0) E) (144, 0)

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To find the point on the curve closest to (0,16), we need to minimize the distance between the two points. Let the point on the curve be (x,y). Then, the distance between (0,16) and (x,y) is given by D^2 = (x-0)^2 + (y-16)^2. We can eliminate the variable D by squaring both sides and substituting the equation of the curve: (x² + y²)^2 = D^2 = (x-0)^2 + (y-16)^2 (x² + y²)^2 = x^2 + (y-16)^2 Expanding both sides and simplifying, we get: (x^2 + y^2) - 32y + 256 = 0 This is the equation of a circle with center (0,16) and radius 16. To find the closest point on the curve to (0,16), we need to find the point where the circle intersects the curve. Substituting y = 144/x into the equation of the circle, we get: x^2 + (144/x - 16)^2 - 256 = 0 Multiplying both sides by x^2 and simplifying, we get: x^4 - 32x^3 + 256x - 2304 = 0 Factoring, we get: (x-12)^2(x^2-8x+192) = 0 The solutions are x = 12, x = 4+4sqrt(2), x = 4-4sqrt(2). Of these, only x = 12 gives a point on the curve closest to (0,16), which is (12, y) where y = 144/12 - 16 = 4. Therefore, the answer is (B) (0,12).

Find the Point on the Curve X2 + Y2 =

Find the Point on the Curve X² + Y² =