Substitute $x=6$ into the circle equation and solve for $y$. You should get two possible $y$-values; then decide which one is greater than $-3$.

The tangent line at a point on a circle is perpendicular to the radius drawn to that point. Once you know the center and the point of tangency, find the slope of the radius, then find the slope of the tangent using negative reciprocals.

After you find the slope of the tangent and the coordinates of the point of tangency, plug them into point-slope form $y-y_1 = m(x-x_1)$ and then match your simplified equation to one of the answer choices.

Type (x-2)^2 + (y+3)^2 = 29 into Desmos to graph the circle. Then type x = 6 to draw the vertical line, and click the two intersection points; note which one has $y > -3$.

Add a point at the center (2,-3) and another at the chosen intersection point (with $y>-3$). Use Desmos’s slope tool (or manually compute) to verify the slope of the segment from the center to this point; this is the radius slope.

Enter each candidate equation from the choices into Desmos, one at a time, along with the circle. For each line, check (a) whether it goes through the correct intersection point with $x=6$ and $y>-3$, and (b) whether it just touches the circle at that one point (tangent) instead of cutting through it (two intersection points).

For the line that appears tangent, add a point on that line at the tangency point and use Desmos to inspect the slopes of the line and the radius. Their slopes should be negative reciprocals (their product is $-1$), confirming it is the correct tangent line.

The circle is

A circle in the form $(x-h)^2 + (y-k)^2 = r^2$ has center $(h,k)$ and radius $r$.

So this circle has:

• Center $(2,-3)$
• Radius $\sqrt{29}$ (though we will not actually need the exact radius value).

We are told the tangent point has $x$-coordinate $6$. Any point on the circle with $x=6$ must satisfy the circle equation.

Substitute $x=6$ into the equation of the circle:

Compute $(6-2)^2$:

Subtract $16$ from both sides:

Now take the square root of both sides:

So the two possible $y$-values are:

• $y = -3 + \sqrt{13}$
• $y = -3 - \sqrt{13}$.

We are told that the point on the circle has a $y$-coordinate greater than $-3$.

Compare the two possible $y$-values:

• $-3 + \sqrt{13}$: here we add a positive number $\sqrt{13} > 0$ to $-3$, so this is greater than $-3$.
• $-3 - \sqrt{13}$: here we subtract a positive number from $-3$, so this is less than $-3$.

Therefore, the tangent point must be

The radius goes from the center $(2,-3)$ to the point of tangency $(6,-3+\sqrt{13})$.

The slope of this radius is

A tangent to a circle is perpendicular to the radius at the point of tangency.

If a line has slope $m$, any line perpendicular to it has slope $-\frac{1}{m}$ (the negative reciprocal).

Here, the radius slope is $m_\text{radius} = \frac{\sqrt{13}}{4}$, so the tangent slope is

Now use point-slope form for the line through $(6,-3+\sqrt{13})$:

Simplify $y - (-3+\sqrt{13})$ to $y + 3 - \sqrt{13}$ and then solve for $y$:

This matches choice D: