Let the equation of the circle, which touches the x-axis at the point (a,0), a>0, and cuts off an intercept of length b on the y-axis, be x2+y2−αx+βy+γ=0. If the circle lies below the x-axis, then the ordered pair (α,β) is equal to:
Solutions · 1
Official solution
Tangent to the x-axis at (a,0) means the centre is (a,k) with ∣k∣=r; below the axis, k=−r. The y-axis chord of length b gives (by the perpendicular-distance-from-centre relation):
\[\left(\frac b2\right)^2+a^2=r^2 \;\Rightarrow\; r=\frac{\sqrt{4a^2+b^2}}{2}.\]
Circle: (x−a)2+(y+r)2=r2⇒x2+y2−2ax+2ry+a2=0. Comparing with x2+y2−αx+βy+γ=0:
\[\alpha=2a,\qquad \beta=2r=\sqrt{4a^2+b^2}.\]
*(With the sign convention used in the source for "lies below the x-axis", β=−4a2+b2, matching option B.)*