x² + y² + 8x - 4y - 16 = 0
x² + y² + 8x - 4y = 16
x² + 8x + y² - 4y = 16

Performing a double completing the square (one for x and one for y),

x² + 8x + 4² + y² - 4y + 2² = 16 + 4² + 2²

Remember to add whatever you introduced on the LHS (left hand side) of the equation to the RHS (right hand side) of the equation as well to maintain equality! We cannot just introduce to the LHS without doing so for the RHS as this imbalances the equation.

(x + 4)² + (y - 2)² = 16 + 4² + 2²
(x + 4)² + (y - 2)² = 16 + 16 + 4
(x + 4)² + (y - 2)² = 36
(x + 4)² + (y - 2)² = 6²

It is clear from here that the centre of the circle (O) has coordinates (-4, 2) while the radius of the same circle is 6 units.

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Lines of the form y = k are horizontal lines (and similarly, lines of the form x = k are vertical lines).

The only times you can obtain horizontal tangents are when the the tangent passes through a turning point of the circle.

The highest point and the lowest point on the circle are the two points in which the tangents drawn through each of them are horizontal. But, it is good to know that the locations of the highest point and the lowest point of the circle are directly above and directly below the centre of the circle. This means that the x-coordinates of the highest point and the lowest point are going to be the same as that of the centre of the circle.

x-coordinate of highest point
= x-coordinate of lowest point
= x-coordinate of the centre of the circle (O)
= -4

y-coordinate of highest point
= y-coordinate of O + radius
= 2 + 6
= 8

y-coordinate of lowest point
= y-coordinate of O - radius
= 2 - 6
= -4

The highest point must be (-4, 8) and the lowest point must be (-4, -4).

The horizontal tangents drawn to each of these points will have equations y = 8 and y = -4 respectively.

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Since the second circle has an area 9 times that of the first circle, the radius of the second circle must be 3 times that of the first circle (because we are increasing BOTH the length and the width of the circle each by a scale factor of 3). You will understand this better after you learn the topic on similar shapes and objects (somewhere within Congruency and Similarity in E Maths).

So, the radius of the second circle must be three times the radius of the first circle, or 18 units.

The centre O remains unchanged at (-4, 2).

So, the equation of the second circle is

(x + 4)² + (y - 2)² = 18²
(x + 4)² + (y - 2)² = 324