chord

**mark** · October 18th 2009, 05:20 AM

the origin O is the midpoint of a chord PQ of the circle $(x - 6)^2 + (y + 3)^2 = 65$ , show the gradient of the chord PQ is 2

could someone please show me how you would figure this out, thanks

**nolanfan** · October 18th 2009, 05:29 AM

Equation of chord of a circle when its midpoint is given is $S_1=S_11$ where S is the equation the circle.

Given equation is
$<br /> x^2 + y^2 - 12x +6y -20 = 0$
Equation of a chord when its midpoint $(x_1,y_1)$
is $xx_1 + yy_1 - 6(x+x_1) + 3(y+y_1) - 20 = x_1^2 + y_1^2 - 12x_1 + 6y_1 -20$
Here midpoint is 0rigin
Therefore
$x(0) + y(0) - 6(x+0) + 3(y +0) - 20 = 0^2 + 0^2 -12(0) + 6(0) - 20$

Therefore equation of chord is
$6x - 3y = 0$ whose slope is -6/-3 =2

**mark** · October 18th 2009, 05:32 AM

i don't think i understand that, is that called differentiation? because i haven't started doing that yet. is there a more simple way to do it?

**nolanfan** · October 18th 2009, 05:56 AM

Okay I will suggest another method.

Let
y= mx be chord to circle
centre is C (6,-3)
Mid pt is T(0,0)

Slope of CT = 0-(-3)/0-6 = -1/2

Since CT is perpendicular to the chord, slope of the chord = m= 2

the equation becomes y = 2x

therefore gradient = 2

**mark** · October 18th 2009, 06:16 AM

i think there's something wrong here, the book asked find the length of the chord PQ and the answer was $4\sqrt{5}$ but the length of the centre to the origin is $3\sqrt{5}$ which i think is what you were talking about. the centre (6, -3) is labelled as C by the way, so i'm not sure if it has anything to do with PQ

**nolanfan** · October 18th 2009, 06:56 AM

Well, I made a mistake by taking the midpoint (0,0) to be P randomly. I've changed the midpoint now to T. I didn't notice that the ends of the chord were P and Q. So let T (0,0) be the midpoint of PQ. Now the length of CT is $3\sqrt5$ . But $\Delta$ CTP is a right angled triangle. therefore,
By pythogarus theorem
$<br /> CP^2 = CT^2 + TP^2$
where TP = 1/2 length of chord.

Also CP = radius = $\sqrt65$

$\sqrt65^2 = 3\sqrt5^2 + TP^2$

$TP^2 = 20$

$TP = 2\sqrt5$

Length of chord = 2 x TP = $4\sqrt5$

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