1.Circle x2+y2−8x=0x^2+y^2-8x=0x2+y2−8x=0 and hyperbola x2/9−y2/4=1x^2/9-y^2/4=1x2/9−y2/4=1 intersect at A,BA,BA,B. Common tangent with positive slope isa.2x−5y−20=02x-\sqrt{5}y-20=02x−5y−20=0b.2x−5y+4=02x-\sqrt{5}y+4=02x−5y+4=0c.3x−4y+8=03x-4y+8=03x−4y+8=0d.4x−3y+4=04x-3y+4=04x−3y+4=0Login to continueOnly logged in users canattempt or see the solution.
