Question 648·Hard·Math

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In the $xy$ -plane, the line with equation $y = mx + 3$ is tangent to the circle $(x - 2)^2 + (y + 1)^2 = 10$ . If $m > 0$ , what is the value of $m$ ?

Your answer

A tangent touches a circle at exactly one point. Substitute the line’s expression for $y$ into the circle equation to get a quadratic in $x$ . Then set its discriminant, the expression $b^2-4ac$ that controls the number of real roots, to zero and solve for the slope with a Desmos regression. Keep the sign restriction in view: another slope may also give a tangent.

Hints

Hint 1
A tangent meets a circle at one point, not two. Replace $y$ in the circle equation with the line’s expression $mx+3$ . How many real $x$ -values should the resulting equation have?
Hint 2
Rearrange the substituted equation into a quadratic in $x$ , meaning an equation whose highest power of $x$ is $x^2$ . Treat $m$ as a coefficient while you collect the $x^2$ , $x$ , and constant terms.
Hint 3
The discriminant $b^2-4ac$ is zero when $ax^2+bx+c=0$ has one real root. Set it to zero for this tangent, then use the given sign restriction when you solve for $m$ .

Step-by-step

Use the one-intersection condition
Step 1
Substitute the line into the circle
A tangent touches the circle at exactly one point. At that point, $y=mx+3$ , so replace $y$ in the circle equation with $mx+3$ :
$(x-2)^2+(mx+4)^2=10.$
This equation must have exactly one real solution for $x$ .
Step 2
Expand both squares
Expand both squares:
$x^2-4x+4+m^2x^2+8mx+16=10.$
Step 3
Collect matching powers of $x$
Collect matching powers of $x$ :
$(1+m^2)x^2+(8m-4)x+20=10.$
Step 4
Set the quadratic equal to zero
Subtract $10$ from both sides:
$(1+m^2)x^2+(8m-4)x+10=0.$
The variable in this quadratic is $x$ ; $m$ is the slope we still need to find.
Step 5
Make the discriminant zero
For a quadratic $ax^2+bx+c=0$ , the discriminant $b^2-4ac$ is zero when there is one real $x$ -value. Here $a=1+m^2$ , $b=8m-4$ , and $c=10$ , so:
$(8m-4)^2-4(1+m^2)(10)=0.$
One tangent point means one root, so set the discriminant to zero.
Step 6
Expand the equation for the slope
Expand, keeping the negative middle term in $(8m-4)^2$ :
$64m^2-64m+16-40-40m^2=0.$
Step 7
Combine like terms
Combine like terms:
$24m^2-64m-24=0.$
Step 8
Reduce the equation
Divide every term by $8$ :
$3m^2-8m-3=0.$
Step 9
Solve for the positive slope
Type $3m^2-8m-3\sim0\{m>0\}$ in Desmos. The $\sim$ asks Desmos to fit $m$ , and $\{m>0\}$ tells it to keep only a positive slope. Under PARAMETERS, Desmos shows $m=3$ . So the tangent line with positive slope has slope $3$ . Grid in 3.

Question 648·Hard·Math

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