Question 18·25 Super-Hard SAT Advanced Math Questions·Advanced Math

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In the $xy$ -plane, the distinct points $P$ and $Q$ are the solutions to the system below, where $k$ is a constant.

\begin{aligned} x^2+y^2-8x+2ky+7&=0 \\[1.4em] y&=x+1 \end{aligned}

Both $P$ and $Q$ have positive $y$ -coordinates, and the distance between $P$ and $Q$ is $2\sqrt{2}$ . What is the value of $k$ ?

Your answer

For a line-circle system with two intersections a fixed distance apart, you don't need both points' coordinates. Substitute the line into the circle to get a quadratic, then use the line's slope to turn the distance into a gap between its roots. A distance-only result can leave more than one constant, so apply any coordinate-sign condition before solving for it in Desmos.

Hints

Hint 1
On the line $y=x+1$ , a horizontal move raises $y$ by the same amount. In the distance formula, what horizontal change makes the point-to-point distance $2\sqrt{2}$ ? Focus on the equal horizontal and vertical changes.
Hint 2
Replace $y$ with $x+1$ in the first equation. The two intersection points then give two roots of one quadratic in $x$ . How can its coefficients tell you the roots' sum and product?
Hint 3
The two $x$ -coordinates have a fixed gap. Their sum lets you find the smaller one, and $y=x+1$ turns it into the smaller $y$ -coordinate. Why must that value be greater than $0$ ?

Step-by-step

Use the distance as a gap between roots
Step 1
Find the horizontal gap
On $y=x+1$ , moving right by $d$ also moves up by $d$ . The distance formula makes the distance between the points $d\sqrt{2}$ . Match it to the given distance, using these equal horizontal and vertical changes:
$d\sqrt{2}=2\sqrt{2}.$
Divide by $\sqrt{2}$ :
$d=2.$
Step 2
Substitute the line into the first equation
At each intersection, $y=x+1$ . Replace $y$ in the first equation so each point's $x$ -coordinate satisfies:
$x^2+(x+1)^2-8x+2k(x+1)+7=0.$
Step 3
Make a quadratic in $x$
Expand the square and distribute $2k$ :
$x^2+x^2+2x+1-8x+2kx+2k+7=0.$
Combine like terms:
$2x^2+(2k-6)x+2k+8=0.$
Divide every term by $2$ :
$x^2+(k-3)x+(k+4)=0.$
Step 4
Read the roots' sum and product
Call the two $x$ -coordinates $x_1<x_2$ . For $x^2+bx+c=0$ , the roots add to $-b$ and multiply to $c$ : expanding $(x-x_1)(x-x_2)$ shows why. Read the sum and product from this quadratic:
$x_1+x_2=3-k,\qquad x_1x_2=k+4.$
Step 5
Use the positive-coordinate condition
The $x$ -coordinates differ by $2$ . Use their sum to find the smaller one:
$x_1=\frac{(3-k)-2}{2}=\frac{1-k}{2}.$
Use $y=x+1$ to find its $y$ -coordinate:
$y_1=\frac{1-k}{2}+1=\frac{3-k}{2}.$
Both $y$ -coordinates are positive only if this smaller one is positive. Multiply the inequality by $2$ :
$3-k>0.$
Rearrange:
$k<3.$
Step 6
Turn the root gap into an equation
The squared gap between two roots equals their squared sum minus four times their product. This follows by expanding both sides:
$(x_2-x_1)^2=(x_1+x_2)^2-4x_1x_2.$
The gap is $2$ , so put the sum and product from above into that identity:
$4=(3-k)^2-4(k+4).$
Step 7
Solve for the allowed constant in Desmos
Type $4\sim(3-k_1)^2-4(k_1+4)\{k_1<3\}$ . This regression asks Desmos to fit the unknown $k_1$ ; the restriction keeps the positive-coordinate condition. Under PARAMETERS, Desmos shows $k_1=-1$ . So the required value of $k$ is $-1$ . Grid in -1.

Question 18·25 Super-Hard SAT Advanced Math Questions·Advanced Math

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