Find the equation to the circle x² + y² - 4x - 2y = 0 at the point (1, 3).

Correct Answer is : 2y - x -5 = 0

Correct Answer is : 2y - x -5 = 0

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WAEC Further Mathematics 2018 Paper

Question : 23

Total: 40

x^{2} + y^{2} - 4x - 2y = 0

at the point (1, 3).

Solution:

Concept:

The equation of the tangent to a circle is found by differentiating the circle equation to get the gradient at the given point.

Explanation:

Given circle:

x^{2} + y^{2} - 4x - 2y = 0

Differentiate implicitly with respect to

x

2x + 2y\frac{dy}{dx} - 4 - 2\frac{dy}{dx} = 0

Group the derivative terms:

(2y - 2)\frac{dy}{dx} = 4 - 2x

\frac{dy}{dx} = \frac{4 - 2x}{2y - 2}

At the point

(1, 3)

\frac{dy}{dx} = \frac{4 - 2(1)}{2(3) - 2} = \frac{2}{4} = \frac{1}{2}

The tangent line through

(1, 3)

with slope

\frac{1}{2}

is:

\frac{y - 3}{x - 1} = \frac{1}{2}

Cross-multiply:

2(y - 3) = 1(x - 1)

2y - 6 = x - 1

Therefore:

2y - x - 5 = 0

Answer:

2y - x - 5 = 0

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