Problem: If the graph of x2+y2=m is tangent to the graph of x+y=2m​, then:
Answer Choices:
A. m must equal 21​
B. m must equal 2​1​
C. m must equal 2​
D. m must equal 2
E. m may be any non-negative real number
Solution:
Method I. Let OA be the x-intercept of the straight line x+y=2m​ and let OB be its y - intercept. Then OA=OB=2m​ so that AOB is a 45∘−45∘−90∘ triangle.
Let the point of tangency be labeled C. Since the graph of x2+y2=m is a circle, OC, the radius r of the circle is perpendicular to AB; that is, OC is the median to hypotenuse AB. Therefore, OC=r=21​⋅2m​=m​, and this is the same value of the radius given by the equation x2+y2=m=(m​)2. Consequently, m may be any non-negative real number.
Method II. Using the distance formula from a line to a point, we have d=A2+B2​∣Ax1​+By1​+C∣​ where Ax+By+C=0 is the given line and ( x1​,y1​ ) is the given point. The given line is x+y−2m​=0 and the given point is (0,0). Therefore, the required d=12+12​∣0+0−2m​∣​=m​. But d=r, the radius of the circle x2+y2=m=(m​)2. Hence, m may be any non-negative real number.
Method III. Let ( x,y ) be the intersection point of the graphs of x2+y2=m and x+y=2m​. This pair of equations yields xy=2m​ and x+y=2m​. Thus x and y may be taken as the roots of t2−2m​t+2m​=0. For tangency the discriminant of this last equation must equal zero. Therefore 2m−4⋅2m​=0, an identify in m. Hence, m may be any non-negative real number.