Circles! oh no!!!

Geometry Level 4

$If\quad a\quad chord\quad of\quad the\quad circle\quad { x }^{ 2 }+{ y }^{ 2 }-4x-2y-c=0\\ is\quad trisected\quad at\quad the\quad points\quad \left( \frac { 1 }{ 3 } ,\frac { 1 }{ 3 } \right) \quad and\quad \\ \left( \frac { 8 }{ 3 } ,\frac { 8 }{ 3 } \right) ,\quad then\quad the\quad radius\quad of\quad the\quad circle\quad will\quad be$

The answer is 5.

3 solutions

Niranjan Khanderia
Mar 16, 2016

{(8/3,8/3) - (1/3, 1/3)} = (7/3, 7/3)= 1/3 * Chord Length.
End points of the chord are, (8/3, 8/3)+(7/3, 7/3)=(5, 5) and (1/3, 1/3) - (7/3, 7/3)=(- 2, - 2).
Circle is (X - 2) $^2$ + (Y - 1) $^2$ = c +20 = r $^2$ .
Using any of the two end points, we get r = 5.

Since I am unable to open reply for Divyam Bapna , request to him to give the detailed steps so all can follow his approach.

Niranjan Khanderia - 5 years, 2 months ago

Divyam Bapna
Jun 9, 2015

Length of the chord is 3 times the distance between the points of trisection. We also know the distance of chord from centre which is 1/√2. Using pythagoras theorem radius = 5

Hrishikesh Dani
May 15, 2015

As the two points lie of the line equation x=y, them the distance between them is equal to the distance between any pint and the point at which the chord has intersected the circle. Thus as dist. between them is 2root2 thus the next point will be {in the positive direction of x-axis} is (5,5) lies on the circle. Thus by putting this value we'll get the value of "-c"= -20. And by using radius formula under root g^2 + f^2 - c =r ( here g=-2, f=1) we will get r=5. Simple...

Circles! oh no!!!

The answer is 5.

3 solutions

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