The equation of a circle to which the line x = 4 is tangent at the point (4, 6) and the line x = 10 is tangent at (10, 6) has to be found.

It can be seen that the lines x = 4 and x = 10 are both...

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The equation of a circle to which the line x = 4 is tangent at the point (4, 6) and the line x = 10 is tangent at (10, 6) has to be found.

It can be seen that the lines x = 4 and x = 10 are both vertical lines and have an equal slope. As they are perpendicular to the circle, the line joining the points of contact (4, 6) and (10, 6) is a diameter of the circle and passes through the center.

The center of the circle is `((4 + 10)/2, (6 +6)/2)` = `(7, 6)`

The radius of the circle is `(1/2)*sqrt((4 - 10)^2 + (6 - 6)^2)` = `(1/2)*sqrt 36` = 3

Usingt the coordinates of the center and the radius the equation of the circle can be written as `(x - 7)^2 + (y - 6)^2 = 9`

**The equation of the required circle is **`(x - 7)^2 + (y - 6)^2 = 9`

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