GCSE Edexcel Maths The Circle: Prove algebraically that the str

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

Question 19

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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 $

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

Find the point of intersection by substituting.

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Answer

To find the point of intersection between the line and the circle, we solve the equations simultaneously. Start by rearranging the equation of the line:

2y = x - 10 \ y = \frac{x - 10}{2}

Next, substitute ( y ) in the circle's equation:

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

Expanding the equation, we get:

x^2 + \frac{(x - 10)^2}{4} = 20

Multiply through by 4 to eliminate the fraction:

4x^2 + (x - 10)^2 = 80

Step 2

Expand and simplify the equation.

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Answer

Now expand ( (x - 10)^2 ):

(x - 10)^2 = x^2 - 20x + 100

Substituting this back into the equation, we get:

4x^2 + x^2 - 20x + 100 = 80 \ 5x^2 - 20x + 100 - 80 = 0 \ 5x^2 - 20x + 20 = 0

Dividing by 5 gives:

x^2 - 4x + 4 = 0

Step 3

Solve the quadratic equation.

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Answer

We factor the quadratic equation:

(x - 2)^2 = 0

The solution is:

x = 2

Substituting ( x = 2 ) back into the equation for ( y ):

y = \frac{2 - 10}{2} = -4

Thus, the point of intersection is ( (2, -4) ).

Step 4

Check the tangency condition.

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Answer

To ensure that the line is a tangent to the circle at ( (2, -4) ), we will calculate the distance from the center of the circle to the line. The center of the circle ( x^2 + y^2 = 20 ) is at ( (0, 0) ) and the radius is ( \sqrt{20} ).

Using the formula for the distance from a point ((x_0, y_0)) to a line (Ax + By + C = 0):

d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}

For the line ( x - 2y - 10 = 0 ) (where ( A = 1, B = -2, C = -10 )) and the point ( (0, 0) ):

d = \frac{|1(0) + (-2)(0) - 10|}{\sqrt{1^2 + (-2)^2}} = \frac{10}{\sqrt{5}} = 2\sqrt{5}

Since this distance ( 2\sqrt{5} ) is equal to the radius ( \sqrt{20} = 2\sqrt{5} ), the line is a tangent to the circle.

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

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

Find the point of intersection by substituting.

Expand and simplify the equation.

Solve the quadratic equation.

Check the tangency condition.

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