Limited Geometry

Geometry Level 3

Find the radius of a circle which passes through the point $(2,0)$ and whose centre is the limit of the point of intersection of the line:

$3x+5y=1$ and $(2+c)x+5{c}^{2}y=1$ as $c\rightarrow 1$ .

The radius of the form $\frac { a\sqrt { b } }{ d }$ where $a$ and $d$ are positive co-prime integers and $b$ is not a perfect square.

Find: $a+b+d$ .

The answer is 1627.

1 solution

Ajit Athle
Jul 4, 2015

Solve the given equations to find the point of intersection as : (2/5,- 1/25) at c=1. Hence our circle is: (x-2/5)²+(y+1/25)²=R². If it has to pass through (2,0) then R=(√1601)/25 giving a+b+c=1627

$R=\frac{\sqrt{1601}}{25}$

This would mean that $a=1,b=1601,c=25$ . The question states that "c is not a perfect square". But, 25 is a perfect square. Please modify the question to avoid confusion.

Janardhanan Sivaramakrishnan - 5 years, 11 months ago

Report resolved

Aditya Kumar - 5 years, 11 months ago

Thanks for posting a solution. Even I did it in the same way. +1

Aditya Kumar - 5 years, 11 months ago

When c=1, the points of intersection are infinite along the line 3x+5y=1 so the point of intersection is (2/5, -1/25) only as c -> 1 not if c=1.

(Note c is representing a different value in the solution to the question which makes it a bit poorly stated at present).

Chris Cooper - 5 years, 11 months ago

Limited Geometry

The answer is 1627.

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