Maths Olympiad Prep

Track / Stage 3 / 183 of 260 #183 of 1964

Problem 183

AMC 10/12, early questions
Geometry Difficulty 3.6 Find the answer

Given a circle (x-a)2+y2=9 (a>5) with a point M on it such that |OM|=2|MQ| (O is the origin) holds true, Q(2,0), the range of values for the real number a is _____.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

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Official solution

Let M(x, y). From |OM|=2|MQ|, we get x2+y2\sqrt {x^{2}+y^{2}}=2(x2)2+y2\sqrt {(x-2)^{2}+y^{2}}, which simplifies to x2+y2-163\frac {16}{3}x+163\frac {16}{3}=0. The center of this circle is (83\frac {8}{3},0) with a radius of 43\frac {4}{3}.

The problem is transformed into finding the intersection of the circle (x-a)2+y2=9 and the circle x2+y2-163\frac {16}{3}x+163\frac {16}{3}=0.

Hence, 3-43\frac {4}{3}≤|a-83\frac {8}{3}|≤3+43\frac {4}{3}, which gives 133\frac {13}{3}≤a≤7. Since a>5, we have 5<a≤7.

Therefore, the answer is 5<a7\boxed{5<a\leq7}.

This problem involves the relationship between a line and a circle and is of medium difficulty.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.