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Geometry Difficulty 3.6 AMC 10/12 Find the answer

If the line 4ax3by+48=04ax-3by+48=0 (a,bRa,b \in \mathbb{R}) always bisects the circumference of the circle x2+y2+6x8y+1=0x^2+y^2+6x-8y+1=0, then the range of values for abab is \_\_\_\_\_\_\_\_.

A number or a short expression. Spacing and $ signs are ignored.

Solution

Analysis

This question mainly examines the positional relationship between a line and a circle, and the application of completing the square, which is a basic problem. According to the problem, the line 4ax3by+48=04ax-3by+48=0 passes through the center of the circle (3,4)(-3,4), hence we have a+b=4a+b=4. Then, using ab=(4b)b=b2+4b=(b2)2+4ab=(4-b)b=-b^2+4b=-(b-2)^2+4, we can find the range of values for abab.

Solution

Given that the line 4ax3by+48=04ax-3by+48=0 bisects the circumference of the circle x2+y2+6x8y+1=0x^2+y^2+6x−8y+1=0,

It follows that the line 4ax3by+48=04ax-3by+48=0 passes through the center of the circle (3,4)(-3,4),

Therefore, we have a+b=4a+b=4,

Thus, ab=(4b)b=b2+4b=(b2)2+44ab=(4-b)b=-b^2+4b=-(b-2)^2+4\leqslant 4,

Therefore, the range of values for abab is (,4](-\infty,4].

Hence, the answer is (,4]\boxed{(-\infty,4]}.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.