After substitution and simplification, you should have a quadratic equation in $x$. Think about what condition must be true for a quadratic to have no real solutions.

Write the discriminant $D = b^2 - 4ac$ in terms of $t$ for your quadratic in $x$, then set up an inequality that reflects the no-real-solution condition and solve that inequality for $t$.

Once you find the range of real $t$ values from the inequality, restrict to nonnegative integers and choose the smallest one that fits.

In Desmos, enter the circle as x^2 + y^2 - 6x - 8y + 9 = 0. Desmos will plot the circle corresponding to the first equation.

Type y = 3x + a so that Desmos creates a slider for a. This slider represents the parameter $t$ in the problem.

Move the slider for a and watch how the line moves relative to the circle. Look for the point where the line just stops intersecting or touching the circle and lies completely outside it. Note the smallest nonnegative value of a at which there are no intersection points; that value of a is the $t$ you are looking for.

We have the system

Substitute $y = 3x + t$ into the first equation:

Expand $(3x+t)^2$ and distribute the $-8$:

• $(3x+t)^2 = 9x^2 + 6tx + t^2$
• $-8(3x+t) = -24x - 8t$

So the equation becomes

Combine like terms:

which we can rewrite as a quadratic in $x$:

The quadratic in $x$ is

where $a = 10$, $b = 6t - 30$, and $c = t^2 - 8t + 9$.

For a quadratic $ax^2 + bx + c = 0$:

• Real solutions occur when the discriminant $D = b^2 - 4ac \ge 0$.
• No real solutions occur when $D < 0$.

So here we need $D < 0$.

Compute the discriminant:

First expand each part:

• $(6t - 30)^2 = 36t^2 - 360t + 900$
• $4(10)(t^2 - 8t + 9) = 40t^2 - 320t + 360$

So

Simplify the expression for $D$:

For no real solutions we need $D < 0$:

Since $-4$ is negative, dividing both sides by $-4$ reverses the inequality:

Now we must solve this inequality in $t$, remembering that $t$ is a nonnegative integer.

First find the roots of the related equation

Use the quadratic formula:

Since $640 = 64 \cdot 10 = 8^2 \cdot 10$, we have $\sqrt{640} = 8\sqrt{10}$, so

Thus the two real roots are $t = -5 - 4\sqrt{10}$ and $t = -5 + 4\sqrt{10}$.

Because the coefficient of $t^2$ is positive, the parabola opens upward, so

for $t$ values outside the interval between the roots:

We are told that $t$ is a nonnegative integer, so only the inequality

can apply.

Now estimate $-5 + 4\sqrt{10}$ to see between which integers it lies. Since $3.1^2 = 9.61$ and $3.2^2 = 10.24$, we have $3.1 < \sqrt{10} < 3.2$, so

Therefore

So $-5 + 4\sqrt{10}$ is between $7$ and $8$.

The inequality $t > -5 + 4\sqrt{10}$ then means $t$ must be larger than a number between $7$ and $8$, and $t$ must be a nonnegative integer. The smallest such integer is

$t = 8$.