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Prove algebraically that the straight line with equation $x - 2y = 10$ is a tangent to the circle with equation $x^2 + y^2 = 20$ - Edexcel - GCSE Maths - Question 19 - 2017 - Paper 3

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Prove-algebraically-that-the-straight-line-with-equation---$x---2y-=-10$---is-a-tangent-to-the-circle-with-equation---$x^2-+-y^2-=-20$-Edexcel-GCSE Maths-Question 19-2017-Paper 3.png

Prove algebraically that the straight line with equation $x - 2y = 10$ is a tangent to the circle with equation $x^2 + y^2 = 20$

Worked Solution & Example Answer:Prove algebraically that the straight line with equation $x - 2y = 10$ is a tangent to the circle with equation $x^2 + y^2 = 20$ - Edexcel - GCSE Maths - Question 19 - 2017 - Paper 3

Step 1

Substitute the equation of the line into the equation of the circle

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Answer

To find the point of intersection between the line and the circle, we first express y in terms of x from the line equation:

2y=x−10⇒y=x−1022y = x - 10 \\ \Rightarrow y = \frac{x - 10}{2}

Now, we substitute this expression for y into the circle's equation:

x^2 + \left( \frac{x - 10}{2} \right)^2 = 20$$

Step 2

Expand and simplify the equation

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Answer

Expanding the substitution gives:

⇒x2+(x−10)24=20⇒4x2+(x−10)2=80⇒4x2+(x2−20x+100)=80⇒5x2−20x+100−80=0⇒5x2−20x+20=0\Rightarrow x^2 + \frac{(x - 10)^2}{4} = 20 \\ \Rightarrow 4x^2 + (x - 10)^2 = 80 \\ \Rightarrow 4x^2 + (x^2 - 20x + 100) = 80 \\ \Rightarrow 5x^2 - 20x + 100 - 80 = 0 \\ \Rightarrow 5x^2 - 20x + 20 = 0

Step 3

Factor or use the quadratic formula to solve for x

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Answer

We simplify the equation:

5(x2−4x+4)=0⇒5(x−2)2=05(x^2 - 4x + 4) = 0 \\ \Rightarrow 5(x-2)^2 = 0

This means:

(x−2)2=0⇒x=2(x - 2)^2 = 0 \\ \Rightarrow x = 2

Step 4

Find the corresponding y value

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Answer

Substituting x = 2 back into the line equation to find y:

2y=2−10⇒2y=−8⇒y=−42y = 2 - 10 \\ \Rightarrow 2y = -8 \\ \Rightarrow y = -4

The point of tangency is (2, -4).

Step 5

Verify the tangency condition

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Answer

Finally, we check that this point (2, -4) lies on the circle:

22+(−4)2=4+16=202^2 + (-4)^2 = 4 + 16 = 20

Since we get the same value as the right side of the circle's equation, we conclude that the line is tangent to the circle.

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