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# If x²+y²=c² and X/a+y/b=1 intersect at A and B then find AB .hence deduce the condition that the line touches the circle

2075 Points
one year ago
hello suman. so the chord intersects circle at A and B. if you drop a perpendicular from the centre O of circle on the chord at P, it will bisect AB (standard theorem).
hence AB= 2PA= 2*sqrt(r^2 – (OP)^2) by pyth theorem, where r is the radius of circle.
note that x²+y²=c² has centre O(0, 0) and radius = c.
so, AB= 2*sqrt(c^2 – (OP)^2).
distance of O(0, 0) from line X/a+y/b=1 or bx + ay – ab= 0 will be:
OP = |b*0 + a*0 – ab|/sqrt(b^2 + a^2)
so, AB= 2*sqrt(c^2 – a^2b^2/(b^2 + a^2)).
now, for tangency AB= 0 as tangent touches circle at one point only.
so, 2*sqrt(c^2 – a^2b^2/(b^2 + a^2))= 0
or c^2(b^2 + a^2) = a^2b^2
or 1/a^2 + 1/b^2 = 1/c^2
kindly approve :))