  Question

# Column 1Column 2a. If ax+by−5=0 is the equation of the chord of the circle    p. 6(x−3)2+(y−4)2=4,which passes through (2,3) and at   the greatest distance from the centre, then |a+b| is equal tob. Let O be the origin and P be a variable point on the circle     q. 3x2+y2+2x+2y=0. If the locus of midpoint of OP is    x2+y2+2gx+2fy+c=0,then(g+f)is equal to            c. The x–coordinate of the centre of the smallest circle which    r. 2   cuts the circles x2+y2−2x−4y−4=0 and x2+y2−10x+12y+52=0 orthogonally is                                                 d. If θ be the angle between two tangents which are drawn        s. 1   to the circlesx2+y2−6√3x−6y+27=0 from the origin, then 2√3tanθ equals to                                                            Which of the following is correct?a→s, b→r, c→p, d→qa→r, b→s, c→p, d→qa→s, b→r, c→q, d→pa→r, b→s, c→q, d→p

Solution

## The correct option is D a→r, b→s, c→q, d→pa. We know that (2,3) lies inside the circle, The chord which is at the greatest distance from the centre is bisected at (2,3),  Slope of line joining point (2,3) and centre is  m=4−33−2=1 The chord will be perpendicular to this, so th slope of the chord =−1 Equation of chord is y−3=−(x−2)⇒x+y−5=0  Comparing with ax+by−5=0, we get a1=b1=−5−5⇒a=b=1∴|a+b|=2 b. Let P be the point (α,β) Assuming the midpoint be M(h,k), so (h,k)=(α2,β2)⇒α=2h,β=2k  Then P=(2h,2k) So putting the point P in the equation of circle, we get 4h2+4k2+4h+4k=0⇒h2+k2+h+k=0 Therefore, the locus of the midpoint is  x2+y2+x+y=0 Comparing with x2+y2+2gx+2fy+c=0, we get f=g=12,c=0∴f+g=1 c. Given circles are  x2+y2−2x−4y−4=0 C1=(1,2),  r1=3 x2+y2−10x+12y+52=0 C2=(5,−6),  r2=3 The equation of line joinig C1C2 is y−2=−6−25−1(x−1)⇒y−2=−2x+2⇒2x+y−4=0⋯(1) The equation of the radical axis is 8x−16y−56=0⇒x−2y−7=0⋯(2) Finding the point of intersection of (1) and (2), we get 5x−15=0⇒x=3 Hence, the required x− coordinate is 3. d. Given circle is x2+y2−6√3x−6y+27=0 C=(3√3,3),  r=3 Distance between centre and origin  d=√27+9=6 Now, we know that sinθ2=rd⇒sinθ2=36=12⇒θ2=π6⇒θ=π3⇒tanθ=√3∴2√3tanθ=6  Suggest corrections   