<p>If three different circles are drawn on a piece of paper, at most how many points can be common to all three? </p>
<p>a)none
b)one
c)two
d)three
e)six</p>
<p>How do you do this type of problem?</p>
<p>I think the answer is C, because two circles will have 2 commen points at most. And if the third circle wants to have commen points with these two at the same time, they must intersect at these 2 points. So I think the answer is C.</p>
<p>I got C. What’s the answer?</p>
<p>C is the answer, anyone have a better explanation?</p>
<p>The answer is C. When 2 circles are drawn on the same plane, there is three options:
- They can have no points in common
- They can be tangent (1 point of intersection)
- They can intersect at 2 points</p>
<p>This can be shown by taking the equation of a circle:
(x^2)+(y^2) = (r^2)</p>
<p>if the equation of circle A is: 9x^2 + 25y^2 = 36
and circle B is: 64(x+2)^2 + 36(y+3)^2 = 25</p>
<p>To find their intersection (if there is any), you would set them equal. As you can already tell, you’ll end up getting a quadratic, which, at most, can only have 2 answers (and thus, at most, 2 points of intersection).</p>
<p>If you add a third circle, it can, at most, intersect with circle A at 2 points and circle B at 2 points. To intersect with both A and B at 2 points, it would have to intersect at circle A and circle B’s points of intersection.</p>
<p>Thus, when 3 circles intersect 2 is the maximum number of points of intersection.</p>