Let the abscissae of the two points P and Q on a circle be the roots of x2−4x−6=0 and the ordinates of P and Q be the roots of y2+2y−7=0. If PQ is a diameter of the circle x2+y2+2ax+2by+c=0, then the value of (a+b−c) is \_\_\_\_\_\_\_\_\_\_.
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Official solution
The circle with diameter endpoints having x-roots of x2−4x−6=0 and y-roots of y2+2y−7=0 is:
\[x^2-4x-6+y^2+2y-7=0 \implies x^2+y^2-4x+2y-13=0.\]
Comparing with x2+y2+2ax+2by+c=0: 2a=−4⟹a=−2; 2b=2⟹b=1; c=−13.
a+b−c=−2+1−(−13)=12.