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Prove that from a point (a,b) of the circle x(x-a) + y(y-b) = 0, two chords, each bisected by the axis of x (rather x-axis), can be drawn if a2>8b2

[This is not my doubt, so provide complete solution for everyone's benefit ]

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12 Answers

62
Lokesh Verma ·

good one.. someone trying?

11
Mani Pal Singh ·

sir question ki language ko ek baar check karna
i wonder some wrong info is given................

1
The Scorpion ·

nahi mani... question's perfectly rite... try karo...

11
Mani Pal Singh ·

Prove that from a point (a,b) of the circle x(x-a) + y(y-b) = 0, two chords, each bisected by the axis of x (rather x-axis), can be drawn if a2>8b2

BOLDED KA KYA MATLAB HAI???

1
°ღ•๓яυΠ·

thr's sum fishy in this q ....................past year iit q hai:P

method goes like this

let it pass thru (h,0) ..........this is d pt wich bisects d chord
so itz equation is

T1=S1

nw it also passes thru (a,b) .accordin 2 ur q

so satisfy dt condition
then u get a qudratic in h

solve for b^2>4ac

1
Ragadeepika ·

Given circle --> x2 + y 2 = ax + by

Let the mid-point of the chord be (p,0)

Equation of chord having mid-point (p,0) is

T = S1

px - a/2(x+p) - b/2y = p2 - ap

2p2 - ap - 2px + ax + by = 0

(a,b) is a point on this chord

2p2 - 3ap + a2 + b2 = 0

This is a quadratic in p.
Since, there are two values of p (2 chords are drawn)
Discriminant > 0

9a2 - 8(a2 + b2) > 0
a2 >8 b2.

Please tell me if I'm wrong.
ALL THE BEST

1
The Race begins... ·

I agree with ragadeepika. :)

Another method :

Let P(a,b) be the given point, even O(0.0) lies on the given circle. Also observe that O,P form the ends of a diameter, so the centre of given circle will be C(a/2,b/2).

Now, the circle with C,P as ends of diameter should intersect X-axis twice so that two chords can be drawn from P which get bisected by axis of x. (observe this carefully, if needed i'll post explanation)

equation of circle with C,P as ends of diameter is : (x-a)(x-a/2)+(y-b)(u-b/2) = 0

substituting y=0 and applying the condition of discriminant > 0, we get a2>8b2

1
skygirl ·

1
skygirl ·

so x1,-b lies on the circle.

thus it shud satisfy the equation x2+y2=xa+yb

=> x12 + b2 = x1.a - b2

=> x12 - x1.a + 2b2 =0

sice x1 should be real... therfor, D> 0 [not equal ofcrs!]

so, a2 > 8b2 . proved :)

1
The Race begins... ·

well, one guy asked me for explanation to my previous post. so this is for him.

Explanation :

1) The perpendicular from centre of a circle to its chord bisects the chord. [observe with reference to CR2(perpendicular to)PQ2 and CR1(perpendicular to)PQ1]

2) Angle formed by a diameter of a circle in the semi-circle is 90o [observe with reference to angle(CR2P) and angle(CR1P)]

hope this clears ur doubt :)

correct me if i am wrong !

1
skygirl ·

no pinks?? ...... so many diff methods!

1
The Race begins... ·

confirmation please.!

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