Persamaan lingkaran dengan pusat (-4,2) dan menyinggung sumbu Y adalah .... A. x^(2)+y^(2)+8x+4y+4=0 B. x^(2)+y^(2)+8x-4y+4=0 C. x^(2)+y^(2)+8x-4y+16=0 D. x^(2)+y^(2)-8x+4y+16=0 E. x^(2)+y^(2)-8x-4y+16=0

B

The question asks for the equation of a circle with a given center and a specific condition (tangency to an axis). The general equation of a circle in Cartesian coordinates is (x - h)^2 + (y - k)^2 = r^2 , where (h, k) is the center of the circle and r is the radius. 1. Identifying the Center: The center of the circle is given as (-4, 2). Therefore, h = -4 and k = 2. 2. Determining the Radius: The circle is tangent to the y-axis. The distance from the center of the circle to the y-axis is the absolute value of the x-coordinate of the center. Thus, the radius r is the absolute value of -4, which is 4. 3. Formulating the Equation: Substituting h = -4, k = 2, and r = 4 into the general equation, we get (x + 4)^2 + (y - 2)^2 = 4^2. 4. Expanding and Simplifying: Expanding this equation, we get x^2 + 8x + 16 + y^2 - 4y + 4 = 16. Simplifying, we get x^2 + y^2 + 8x - 4y + 4 = 0. 5. Matching with the Options: The equation x^2 + y^2 + 8x - 4y + 4 = 0 matches with option B. Therefore, the correct answer is option B.