The coordinates of the centre of a circle are $(x - 7, 2x)$ . Find the value(s) of 'x', if the circle passes through the point $(-9, 11)$ and has radius $5\sqrt{2}$ units.

Q: 23. The coordinates of the centre of a circle are $(x - 7, 2x)$. Find the value(s) of 'x', if the circle passes through the point $(-9, 11)$ and has radius $5\sqrt{2}$ units.

$x = 3$ or $x = 5$. Setting the distance from centre $(x - 7, 2x)$ to point $(-9, 11)$ equal to the radius $5\sqrt{2}$ and squaring gives $(x + 2)^2 + (2x - 11)^2 = 50$, which simplifies to $x^2 - 8x + 15 = 0$, factorising as $(x - 3)(x - 5) = 0$.

Q: Why are there two values of x?

The condition produces the quadratic $x^2 - 8x + 15 = 0$, which factorises as $(x - 3)(x - 5) = 0$. Both roots give valid centres, $(-4, 6)$ and $(-2, 10)$, each exactly $5\sqrt{2}$ from $(-9, 11)$. The word 'value(s)' signals both are expected.

Q: What is a common mistake in this problem?

The biggest slip is the double negative in $(x - 7) - (-9)$: it equals $x + 2$, not $x - 16$. The second is forgetting that $(5\sqrt{2})^2 = 50$, not $5\sqrt{2}$ or 10.

Coordinate Geometry

Distance formula / equation of circle

Official Answer

$x = 3$ or $x = 5$ . Setting the distance from centre $(x - 7, 2x)$ to point $(-9, 11)$ equal to the radius $5\sqrt{2}$ and squaring gives $(x + 2)^2 + (2x - 11)^2 = 50$ , which simplifies to $x^2 - 8x + 15 = 0$ , factorising as $(x - 3)(x - 5) = 0$ .

distance formula equals radiuscircle centre and radius(5√2)² = 50x² − 8x + 15 = 0(x−3)(x−5)=0x = 3 or x = 5two values of x

Marking Scheme

1½ mark: Setting up distance = radius: $(x + 2)^2 + (2x - 11)^2 = (5\sqrt{2})^2 = 50$ .
2½ mark: Correct expansion and simplification to $5x^2 - 40x + 75 = 0$ (or $x^2 - 8x + 15 = 0$ ).
3½ mark: Correct factorisation $(x - 3)(x - 5) = 0$ .
4½ mark: Both values $x = 3$ and $x = 5$ stated (dropping one root loses this half-mark).

Hint

Distance from the centre to the given point equals the radius. Write $CP^2 = (5\sqrt{2})^2 = 50$ , and remember $(x - 7) - (-9) = x + 2$ . Expect TWO values of x.

Quick Oral Answer

Since the point $(-9, 11)$ lies on the circle, its distance from the centre $(x - 7, 2x)$ equals the radius $5\sqrt{2}$ ; squaring gives $(x + 2)^2 + (2x - 11)^2 = 50$ , which simplifies to $x^2 - 8x + 15 = 0$ , so $x = 3$ or $x = 5$ , both giving valid centres.

Analysis & Explanation

This question links the distance formula to the geometric definition of a circle (every point on it is equidistant from the centre).

Concept

Point $P(-9, 11)$ lies on the circle with centre $C(x - 7, 2x)$ , so distance CP equals the radius $5\sqrt{2}$ .
Squaring removes the surd immediately: $(5\sqrt{2})^2 = 50$ .

Key points

Setting up $(x + 2)^2 + (2x - 11)^2 = 50$ and simplifying gives the clean quadratic $x^2 - 8x + 15 = 0$ , which factorises as $(x - 3)(x - 5) = 0$ .
Both roots $x = 3$ and $x = 5$ are valid — the phrase 'value(s)' signals more than one answer is expected.

Common mistakes

Sign slip in $(x - 7) - (-9) = x + 2$ (double negative).
Discarding one of the two valid roots instead of reporting both.

Common Mistakes

1Sign error in $(x - 7) - (-9)$ : writing $(x - 16)$ instead of $(x + 2)$ by mishandling the double negative.
2Forgetting to square the radius correctly — using $(5\sqrt{2})^2 = 10$ or $5\sqrt{2}$ instead of 50.
3Reporting only one root $(x = 3 \text{ OR } x = 5)$ when the equation clearly yields both, ignoring the plural 'value(s)' in the question.

Interesting Facts

Two distinct centres satisfy the condition because a point at a fixed distance from $(-9, 11)$ traces a circle of radius $5\sqrt{2}$ , and the locus of the moving centre $(x - 7, 2x)$ is a straight line — a line generally cuts a circle in two points, hence two answers.

The distance formula is simply the Pythagorean theorem applied to the horizontal and vertical gaps between two points, first cast in coordinate form by René Descartes in the 17th century.

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Frequently Asked Questions

How is the distance formula used here?

The point $(-9, 11)$ is on the circle, so its distance from the centre $(x - 7, 2x)$ equals the radius $5\sqrt{2}$ . Writing this distance squared equal to $(5\sqrt{2})^2 = 50$ gives $(x + 2)^2 + (2x - 11)^2 = 50$ .

Why are there two values of x?

The condition produces the quadratic $x^2 - 8x + 15 = 0$ , which factorises as $(x - 3)(x - 5) = 0$ . Both roots give valid centres, $(-4, 6)$ and $(-2, 10)$ , each exactly $5\sqrt{2}$ from $(-9, 11)$ . The word 'value(s)' signals both are expected.

What is a common mistake in this problem?

The biggest slip is the double negative in $(x - 7) - (-9)$ : it equals $x + 2$ , not $x - 16$ . The second is forgetting that $(5\sqrt{2})^2 = 50$ , not $5\sqrt{2}$ or 10.