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