The equation of a circle is $(x-a)^2+(y-3)^2=20$

Expert intro: This is a tangent-to-circle problem. The key idea is to substitute the line into the circle, then use the fact that a tangent gives exactly one point of intersection, so the resulting quadratic must have discriminant $0$.

Detailed walkthrough

Let the circle be

$(x-a)^2+(y-3)^2=20$

and the tangent line be

$y=\frac12 x+6.$

1) Substitute the line into the circle

Replace $y$ by $\frac12 x+6$:

$(x-a)^2+\left(\frac12 x+6-3\right)^2=20$

so

$(x-a)^2+\left(\frac12 x+3\right)^2=20.$

Expand:

$x^2-2ax+a^2+\frac14 x^2+3x+9=20$

$\frac54 x^2+(3-2a)x+(a^2-11)=0.$

2) Use the tangent condition

Because the line is tangent to the circle, this quadratic in $x$ has one repeated root, so

$b^2-4ac=0.$

Here

$a_q=\frac54,\quad b_q=3-2a,\quad c_q=a^2-11.$

Hence

$(3-2a)^2-4\left(\frac54\right)(a^2-11)=0.$

Simplify:

$9-12a+4a^2-5a^2+55=0$

$-a^2-12a+64=0$

$a^2+12a-64=0.$

3) Solve for $a$

$a=\frac{-12\pm\sqrt{12^2-4(1)(-64)}}{2}$

$a=\frac{-12\pm\sqrt{144+256}}{2}$

$a=\frac{-12\pm 20}{2}.$

So the two possible values are

$a=4 \quad \text{or} \quad a=-16.$

💡 Pitfall guide

A common mistake is to set the substituted equation equal to $0$ and stop there. For a tangent, the important step is not just forming the quadratic, but forcing its discriminant to be $0$. Also, keep the expansion of $\left(\frac12 x+3\right)^2$ careful: the middle term is $3x$, not $6x$.

🔄 Real-world variant

If the line were not tangent but secant, then the substituted quadratic would have two distinct real roots, so the discriminant would be positive. If the line were parallel to the tangent but shifted, the discriminant could become negative, meaning no intersection. The same substitution method still works for any line $y=mx+c$; only the coefficients change.

🔍 Related terms

tangent to a circle, discriminant, substitution method