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Question

Find the co-ordinates of the middle point of the chord which the circle x2+y22x2=0 cuts off on the line y=x1

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Solution

We have,

Equation of line y=x1 …… (1)

Equation of circle x2+y22x2=0 …… (2)

Substitute the value of y by equation (1) in equation (2) and we get,

x2+y22x2=0

x2+(x1)22x2=0

x2+x2+12x2=0

2x22x1=0

Comparing that ax2+bx+c=0 and we get,

a=2,b=2,c=1

Then we know that,

x=b±b24ac2a

=(2)±(2)24×2×(1)2×2

=2±124

=2±234

=1±32

Then, x1=1+32 and x2=132

Put the value of x in equation (1) and we get,

y=1±321

y=1±322

y=1±32

Then, y1=1+32 and y2=132

Let the middle point of chord.

(x,y)=(x1+x22,y1+y22)

(x,y)=⎜ ⎜ ⎜ ⎜1+32+1322,1+32+1322⎟ ⎟ ⎟ ⎟

(x,y)=⎜ ⎜ ⎜ ⎜1+3+1322,1+31322⎟ ⎟ ⎟ ⎟

(x,y)=⎜ ⎜ ⎜222,222⎟ ⎟ ⎟

(x,y)=(12,12)

Hence, this is the answer.

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