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: 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 $x = 5$, obtained by equating the distance from centre $(x-7, 2x)$ to the point $(-9, 11)$ with the radius $5\sqrt{2}$, giving $x^2 - 8x + 15 = 0$.

Coordinate Geometry

Distance Formula and Circle

Official Answer

$x = 3$ or $x = 5$ , obtained by equating the distance from centre $(x-7, 2x)$ to the point $(-9, 11)$ with the radius $5\sqrt{2}$ , giving $x^2 - 8x + 15 = 0$ .

distance formularadiuscentre of circlequadratic equationcoordinate geometryfactorisation

Marking Scheme

11 mark: correctly setting up the distance equation $\sqrt{(x+2)^2 + (2x-11)^2} = 5\sqrt{2}$ and simplifying to the quadratic $x^2 - 8x + 15 = 0$ .
21 mark: correctly factorising/solving to get both values $x = 3$ and $x = 5$ .

Hint

Use the distance formula between the centre $(x-7, 2x)$ and the given point $(-9, 11)$ , and equate it to the radius $5\sqrt{2}$ .

Quick Oral Answer

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

Analysis & Explanation

Combines the Distance Formula with the definition of a circle (all points equidistant from the centre).

Concept

Distance from centre $(x-7, 2x)$ to a point $(-9, 11)$ on the circle equals the radius $5\sqrt{2}$ .

Key points

Setting up $\sqrt{(x+2)^2 + (2x-11)^2}$ = $5\sqrt{2}$ and squaring leads to the quadratic $x^2 - 8x + 15 = 0$ .
Factorising gives $x = 3$ or $x = 5$ — both valid since no constraint rules either out.

Common mistakes

Sign error simplifying $(x-7-(-9))$ to (x+2).
Arithmetic slip expanding (2x-11)².

Common Mistakes

1Sign error in computing $x - 7 - (-9)$ , incorrectly simplifying it to $x - 16$ instead of the correct $x + 2$ .
2Squaring 5√2 incorrectly as 25√2 or 10 instead of the correct 50.
3Stopping after finding one root of the factorised quadratic and forgetting that both $x = 3$ and $x = 5$ are valid solutions.

Interesting Facts

The distance formula $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$ used here is a direct application of the Pythagoras theorem to coordinate points, one of the most tested formulas across all Class 10 coordinate geometry problems.

A quadratic equation in coordinate geometry problems like this typically yields two valid answers precisely because a circle's centre condition is symmetric — reflecting the geometric fact that many points can be equidistant ( $5\sqrt{2}$ units) from $(-9, 11)$ .

$5\sqrt{2}$ as a radius often arises from a right-triangle-like distance calculation (e.g., legs 5 and 5), since $5\sqrt{2}$ = $\sqrt{50} = \sqrt{25+25}$ .

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Frequently Asked Questions

Why does this question give two values of x?

Because the distance equation, after squaring, becomes a quadratic equation in x, and both roots ( $x=3$ and $x=5$ ) satisfy the original condition without any geometric restriction ruling one out.

What formula connects the centre, a point on the circle, and the radius?

The distance formula: if $(h,k)$ is the centre and $(x_1,y_1)$ lies on the circle with radius r, then $\sqrt{(x_1-h)^2 + (y_1-k)^2} = r$ .