Book's answer is wrong......

Find the condition that the straight line cx-by+b^{2}=0 may touch the circle x^{2}+y^{2}=ax+by and find the point of contact.

3 Answers

1
Arshad ~Died~ ·

havent u heard of the condition for a line to be tangent to a circle???
c=±a√(1+m2)

1
yagyadutt Mishra ·

Hello.....arshad,,,,the condition given by you is only for the standard circle...x^2 + y^2 = a

Now the best method.....is to find the perpendicular distance from centre to the lline and equate with radius...let me try the question..

circle is x^2 + y^2 -ax -by =0
centre is ( a/2,b/2)
radius is r = 1/2*√a^2 + b^2

Now line is cx -by + b^2 = 0
Perpendicular distance = ac/2 - b^2/2 + b^2 /√c^2 + b^2

now equate both and get the relation

1
Riju ·

Yeh got it man..............its more easier by equating the radius after finding the perpendicular distance from the centre on the tangent which infact is the radius as shown by yagyadutt............Thanks

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