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The locus of centre of circle which cuts a chord of length 2a from the positive x axis and passes through a point on positive y axis distance b from the origin is? Ans is x^2+2by=a^2-b^2 but how?

The locus of centre of circle which cuts a chord of length 2a from the positive x axis and passes through a point on positive y axis distance b from the origin is?
Ans is x^2+2by=a^2-b^2 but how?

Grade:12

3 Answers

Prajwal
28 Points
4 years ago
We know length of intercept made by circle on X axis is √(-C )= 2A 
AND.  Length of intercept made by circle on y axis is √(F 2 --c)=b 
Just do the squaring and add them you will get the same equation
Prajwal
28 Points
4 years ago
Correction the legenth of intercept are 2√g^2-c
And 2√f^2-c it is not same as given above  so you will get g^2-c=a^2 and f^2-c=b^2/2 
 
SURYANSH SINGH RAWAT
11 Points
4 years ago
=>2✓g²-c = 2a => g²-c = a².........(1)Now let the equation of circle be x²+y²+2gx+2fy+c=0 It passes through (0,b) so,=> b²+2fb+c=0 .......(2)Here (-g,-f) is (h,k) By solving (1) & (2) by eliminating `c` we get the required equation of the ellipse i.e [x²-2yb+b²=a²]

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