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This question has me stuck at the very beginning and I dont understand what to do. Dont need the solution, just a hint on what to do. Q.A and B are points in the xy plane, which are 2sqrt2 units apart and subtend an angle of 90(degree) at a point C(1,2) on line x - y + 1 = 0 , which is larger than any angle subtended by line segment AB at any other point on the line. The equation(s) of the circle through points A,B and C is/are?

I really dont understand how to solve it...

Jyotinder Singh
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  • I think you might need to rephrase the question to make it a bit clearer – David Quinn May 26 '15 at 14:37
  • Hmm...the question is exactly the same as I have quoted it. Thats the problem, I am unable to understand it... – Jyotinder Singh May 26 '15 at 14:43
  • Hint. $ACB$ is a right triangle with right angle at $C$ and hypotenuse $2\sqrt{2}$. The rest of the data tell you something more about that triangle. Can you draw a picture? – Ethan Bolker May 26 '15 at 14:48

2 Answers2

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Consider a right $\triangle ABC$, with hypotenuse $AB,\ |AB|=2\sqrt2$ and a known location of point $C=(1,2)$.

It is known that for a given hypotenuse $AB$ point $C$ is located on a circle with the centre $O=(A+B)/2$ and radius $r=|AB|/2$, thus the middle point $O$ can be located only on the circle with radius $r$ around the point $C$.

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In case when the line $L$ crosses the hypotenuse $AB$, it is always possible to find another point $C'$ on the line $L$ such that $\angle AC'B>\angle ACB$:

enter image description here

Since $\angle ACB$ is maximal angle, $AB$ should not cross the line $L$.

When we select a point $O$ on the circle $\mathcal{P}$, points $A,B$ should be on the circle $\mathcal{Q}$ with the same radius $r$, and if the line $L$ crosses the circle $\mathcal{Q}$, any point $C'$ on the line inside $\mathcal{Q}$ will provide $\angle AC'B>\angle ACB$:

enter image description here

So, we have to choose point $O$ such that the circle $\mathcal{Q}$ is tangent to the line $L$:

enter image description here

For pair of endpoints of any diameter of $\mathcal{Q}$ could be $A,B$, since they provide the largest $\angle ACB=90$; for any other point $C'$ on the line $\angle AC'B<\angle ACB$.

Note that the question is not to find the points $A,B$, but only the equation(s) of the circle through points $A,B$ and $C$. So, one such a circle is $\mathcal{Q}$, with centre $O=(2,1)$ and radius $r=\sqrt2$ and another is symmetric with respect to the line $L$ with the centre $O'=(0,3)$.

g.kov
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If I understand the question correctly, I think it means that $AB$ must be parallel to the line $x-y+1=0$, and that $AB$ Is a diameter of the circle and that $ABC$ is a right-angled isosceles triangle with the right-angle at $C$. In which case $AC=BC=2$ and the centre of the circle is the midpoint of $AB$, i.e. $(2, 1)$. Can you take it from there?

David Quinn
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