Coordinate Geometry: Question 10

Worked solution

Part (a): Substituting the line into the circle

The circle’s equation is $x^2 + y^2 = 8$. Substitute $y = x + k$:

$x^2 + (x + k)^2 = 8.$

Expand $(x+k)^2 = x^2 + 2kx + k^2$:

$x^2 + x^2 + 2kx + k^2 = 8$

$2x^2 + 2kx + k^2 = 8$

$2x^2 + 2kx + (k^2 - 8) = 0. \quad \blacksquare$

This matches the quadratic given in the question.

Part (b): Finding $k$ using the discriminant

A line is a tangent to a curve exactly when the resulting quadratic has a repeated root, i.e. discriminant $b^2 - 4ac = 0$.

For $2x^2 + 2kx + (k^2 - 8) = 0$, we have $a = 2$, $b = 2k$, $c = k^2 - 8$.

$b^2 - 4ac = (2k)^2 - 4(2)(k^2 - 8) = 4k^2 - 8(k^2 - 8) = 4k^2 - 8k^2 + 64 = -4k^2 + 64.$

Setting this equal to zero for tangency:

$-4k^2 + 64 = 0 \implies 4k^2 = 64 \implies k^2 = 16 \implies k = \pm 4.$

So the two possible values are $\boxed{k = 4 \text{ or } k = -4}$.

Check ($k=4$): $2x^2+8x+(16-8)=2x^2+8x+8=0 \implies x^2+4x+4=0 \implies (x+2)^2=0$, a repeated root. $\checkmark$

Check ($k=-4$): $2x^2-8x+(16-8)=2x^2-8x+8=0 \implies x^2-4x+4=0 \implies (x-2)^2=0$, also a repeated root. $\checkmark$

Part (c): Point of tangency for the positive value of $k$

Using $k = 4$, the quadratic from part (a) becomes:

$2x^2 + 2(4)x + (4^2 - 8) = 0 \implies 2x^2 + 8x + 8 = 0.$

Dividing by $2$:

$x^2 + 4x + 4 = 0 \implies (x + 2)^2 = 0 \implies x = -2 \text{ (repeated root)}.$

Substituting into the line $y = x + k = x + 4$:

$y = -2 + 4 = 2.$

Check on the circle: $(-2)^2 + 2^2 = 4 + 4 = 8. \checkmark$ This matches the circle’s equation $x^2+y^2=8$ exactly.

So for $k = 4$, the line touches the circle at $\boxed{(-2, 2)}$.

Final answers

• (a) $2x^2 + 2kx + (k^2 - 8) = 0$, shown
• (b) $k = 4$ or $k = -4$
• (c) Point of tangency (for $k=4$): $(-2, 2)$