Question

In the standard x,y plane, a circle has a radius 6 and center (7, 3). the circle intersects the x-axis at (a, 0) and (b, 0). what is the value of a + b?

Answer 1

The answer is 14.
Enjoy!

Answer 2

Answer :-

The required value of a+b is 14.

Step-by-step explanation :-

☆ Given :-

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).

☆ 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

$\sf\implies\:(x-h)^2+(y-k)^2=r^2~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(i)$

For the given circle, we have

(h, k) = (7, 3)  and  r = 6 units.

So, from equation (i), we get

$\sf\implies\:(x-7)^2+(y-3)^2=6^2\\\\\Rightarrow \sf\:(x-7)^2+(y-3)^2=36~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(ii)$

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

$\sf\implies\:(c-7)^2+(0-3)^2=36\\\\\Rightarrow (c-7)^2+9=36\\\\\Rightarrow (c-7)^2=27\\\\\Rightarrow c-7=\pm3\sqrt3~~~~~~~~~~~~~~~~~~~~~~[\textup{Taking square root on both sides}]\\\\\Rightarrow c=7\pm3\sqrt3$

Therefore, we get

$\sf\implies\:a=7+3\sqrt3,\\\\b=7-3\sqrt3.$

That is,

$\sf\implies\: a+b=(7+3\sqrt3)+(7-3\sqrt3)=14.$

Thus, the required value of a+b is 14.