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find the radius of the circle with center at (-1,2) if a chord of length 10 is bisected at (4,-3).(this is exactly what our professor given to us)

im thinking of using the distance formula which is

d=√(x2-x1)^2+(y2-y1)^2

but our topic is about division of line segment but i think i cant use that because the problem is about finding the length of radius of a circle link of the formula

and ive tried to graph it, but i dont know if it is correct.

my graph

this is according on my understanding

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  • $\begingroup$ See proofwiki.org/wiki/… $\endgroup$ Jan 15, 2016 at 17:51
  • $\begingroup$ @labbhattacharjee sorry i didn't understand the link :( $\endgroup$
    – leeeeee
    Jan 15, 2016 at 17:56
  • $\begingroup$ I am surprised that the problem has a unique solution. Bisection of a chord seems to be a rather strong property of a given line through a circle. $\endgroup$
    – Imago
    Jan 15, 2016 at 18:03

2 Answers 2

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As the perpendicular bisector of any chord of any given circle must pass through the center O$(-1,2)$ of that circle.

If $P(4,-3)$ is midpoint of the chord and $R$ is one of the extreme points,

$OR^2=OP^2+PR^2, PR=5$ unit

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Let the center be O. It says the chord of length 10 is bisected at (4,-3). Let the chord be AB and its mid point be D which is (4,-3). Now we know, the perpendicular from center to a chord bisects it, so we have two right angular triangles, ∆OAD and ∆OBD where OA will be radius, r. You can calculate OD by using the distance formula. And we know that OD bisects AB so AD = DB = 10.2 =5. So now you simply have to use pythagoras theoram on, say OAD, where OA = r, AD = 5, and OD = 5√2 (you can calculate using distance formula as you have the coordinates of both O and D). So, OA^2 = OD^2 + AD^2. You should get radius as 5√3.

Hope it helps.

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  • $\begingroup$ That's much better. $\endgroup$
    – David K
    Jan 15, 2016 at 19:54

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