Try substituting $y = 2x - 3$ into the circle equation to get an equation with only $x$. What type of equation in $x$ will that be?

Once you have a quadratic equation in $x$, remember that the number of real solutions depends on the discriminant $b^2 - 4ac$. What condition on the discriminant gives exactly one real solution?

After setting the discriminant equal to zero, carefully solve the resulting linear equation for $m$, watching your arithmetic with negatives and fractions.

In Desmos, type m = 0 to create a slider for $m$. Then graph the circle as x^2 + y^2 - 4x + 6y = m and the line as y = 2x - 3. You will see the line and a circle whose size changes when you move the $m$ slider.

Slowly drag the $m$ slider left and right. Watch how the number of intersection points between the line and the circle changes: sometimes 0, sometimes 2. Look for the exact $m$ value at which the line just touches the circle at a single point (the graphs are tangent).

When you find the position of the slider where the line and circle meet at exactly one point, note the corresponding value of $m$ shown by the slider; that is the value of $m$ that makes the system have exactly one real solution.

The first equation is a circle and the second is a line.

• A line and a circle can intersect in 0, 1, or 2 points.
• "Exactly one real solution" means the line just touches the circle (is tangent) or hits a circle that has collapsed to a single point.

Algebraically, after substituting $y=2x-3$ into the circle equation, you will get a quadratic in $x$. For the system to have exactly one real solution, that quadratic must have exactly one real root, which happens when its discriminant is zero.

Substitute $y=2x-3$ into $x^2 + y^2 - 4x + 6y = m$:

Now expand and simplify:

• $(2x - 3)^2 = 4x^2 - 12x + 9$

So the left-hand side becomes

Combine like terms:

• $x^2 + 4x^2 = 5x^2$
• $-12x - 4x + 12x = -4x$
• $9 - 18 = -9$

So we get

Move $m$ to the left to write this as a quadratic in $x$:

This has the form $ax^2 + bx + c = 0$ with $a = 5$, $b = -4$, and $c = -(9 + m)$.

For a quadratic $ax^2 + bx + c = 0$ to have exactly one real solution, its discriminant must be zero:

Here $a = 5$, $b = -4$, $c = -(9 + m)$, so

Simplify step by step:

Solve for $m$:

So, for the system to have exactly one real solution, $m$ must be $-\dfrac{49}{5}$.