Answer: The required value of a+b is 14.
Step-by-step explanation: Given that in the standard xy plane, a circle has a radius 6 and center (7, 3).
The circle intersects the x-axis at (a, 0) and (b, 0).
We are to find the value of a+b.
We know that
the standard equation of a circle with center at (h, k) and radius r units is given by
[tex](x-h)^2+(y-k)^2=r^2~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)[/tex]
For the given circle, we have
(h, k) = (7, 3) and r = 6 units.
So, from equation (i), we get
[tex](x-7)^2+(y-3)^2=6^2\\\\\Rightarrow (x-7)^2+(y-3)^2=36~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)[/tex]
Since the circle (ii) passes through the points (a, 0) and (b, 0), so let the point be denoted by (c, 0), then we have
[tex](c-7)^2+(0-3)^2=36\\\\\Rightarrow (c-7)^2+9=36\\\\\Rightarrow (c-7)^2=27\\\\\Rightarrow c-7=\pm3\sqrt3~~~~~~~~~~~~~~~~~~~~~~[\textup{Taking square root on both sides}]\\\\\Rightarrow c=7\pm3\sqrt3[/tex]
Therefore, we get
[tex]a=7+3\sqrt3,\\\\b=7-3\sqrt3.[/tex]
That is,
[tex]a+b=(7+3\sqrt3)+(7-3\sqrt3)=14.[/tex]
Thus, the required value of a+b is 14.
