A chord of the circle x^2 + y^2 - 4x - 6y = 0 passing through the origin subtends an angle tan^-1 (7/4) at the point where the circle meets positive y axis. equation of the chord is ____ Share with your friends Share 4 Manbar Singh answered this The given equation of the circle is,x2 + y2 - 4x - 6y = 0Put x = 0, we get y2 - 6y = 0⇒yy - 6 = 0⇒y = 0 or y = 6So, the given circle passes through the origin O0,0 and meets poisitive y-axis at B0,6.Let OP be the chord of the circle passing through the origin subtending an angle θ at B, where tan θ = 74So, ∠OBP = θNow, the given equation of the circle is,x2 + y2 - 4x - 6y = 0Comparing the above equation with x2+y2+2gx+2fy + c = 0, we get2g = -4 ⇒ g = -22f = -6 ⇒f = -3Now, coordinates of the centre of the circle are -g,-f =C2,3Slope of CO = 3 - 02 - 0 = 32Slope of the line ⊥ to CO = -1Slope of CO = -23Since CO is a radius and OT is a tangent and we know that tangent to a circle is ⊥ to its radius at the point of contact.So, slope of the tangent OT = -23Equation of the tangent OT passing through O0,0 isy - 0 = -23x - 0⇒2x + 3y = 0Let ∠XOT = αNow, tan α = 23Now, ∠POT =∠OBP = θ Angles in the alternate segmentNow, ∠POX = ∠POT -∠XOT = θ - αNow, tanθ-α = tan θ - tan α1 + tan θ . tan α=74-231+74×23=12So, slope of OP = tanθ-α = 12Now, equation of chord OP passing through 0,0 and having slope 12 is y - 0 = 12x - 0⇒x - 2y = 0 13 View Full Answer