CircleHardBloom L4
Question
A circle touches the x-axis and also touches the circle with center at $(0, 3)$ and radius $2$. The locus of the center of the circle is
Options
A.an ellipse
B.a circle
C.a hyperbola
D.a parabola
Solution
{"given":"A variable circle touches the x-axis and also touches externally a fixed circle with center at (0, 3) and radius 2. We need to find the locus of the center of the variable circle.","key_observation":"Since the variable circle touches the x-axis, its center is at distance equal to its radius from the x-axis. If the center is at (h, k), then the radius is k. For two circles touching externally, the distance between their centers equals the sum of their radii. This constraint will give us the locus equation.","option_analysis":[{"label":"(A)","text":"an ellipse","verdict":"incorrect","explanation":"An ellipse would result from a constraint involving sum of distances to two fixed points, which is not the case here with the tangency conditions."},{"label":"(B)","text":"a circle","verdict":"incorrect","explanation":"A circle would result from a constraint involving constant distance from a fixed point, but our tangency conditions create a different relationship."},{"label":"(C)","text":"a hyperbola","verdict":"incorrect","explanation":"A hyperbola would result from a constraint involving difference of distances to two fixed points, which doesn't match our tangency conditions."},{"label":"(D)","text":"a parabola","verdict":"correct","explanation":"Let center be (h, k). Since it touches x-axis, radius = k. Distance between centers = k + 2 (external tangency). So $\\sqrt{h^2 + (k-3)^2} = k + 2$. Squaring and simplifying gives $h^2 = 10k - 5 = 10(k - \\frac{1}{2})$, which is the standard form of a parabola."}],"answer":"(D)","formula_steps":[]}
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