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Trignometry
From a point P (x1 , y1 ), tangents PA and PB are drawn to the circle x2 +y2 +2gx +2fy +c = 0
If P' and Q' are mid points of PA and PB respectively, find the equation of P'Q'.
Comments (17)
x(x'+g)+y(y'+f)+gx'+fy'+c=0
x(x'+g)+y(y'+f)+gx'+fy'+c=0
distance of chord of contact to point P. from similiarity of triangle we will find that the
distance of P'Q' from P will be half of
distance of chord of contact to point P. so distance from P'Q' to P can be known. now we know slope of P'Q' & distance between both parrallel linesPreparing for JEE?
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1)finding the equation of AB
2)finding M(the intersection point of AB and OP)
3)finding the mid point of PM
4)finding the equation of P'Q'
every step is a short step.
in the first step, let A be (x,y) and C be centre of OP. AC=CP completes first step
infourth step, slope and a point are known.
answer comes out to be x(x1+g)+y(y1+f)=PA/2