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A circle has equation $x^2 + y^2 = 12.25$ - Edexcel - GCSE Maths - Question 24 - 2022 - Paper 2

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A circle has equation $x^2 + y^2 = 12.25$. The point P lies on the circle. The coordinates of P are (2.1, 2.8). The line L is the tangent to the circle at point P.... show full transcript

Worked Solution & Example Answer:A circle has equation $x^2 + y^2 = 12.25$ - Edexcel - GCSE Maths - Question 24 - 2022 - Paper 2

Step 1

Find the center and radius of the circle

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Answer

The equation of the circle is given by x2+y2=12.25x^2 + y^2 = 12.25. This is in the standard form (x−h)2+(y−k)2=r2(x - h)^2 + (y - k)^2 = r^2, where (h,k)(h, k) is the center and rr is the radius.

From the equation, we see that the center is at (0,0)(0, 0) and the radius is: r=extsqrt(12.25)=3.5.r = ext{sqrt}(12.25) = 3.5.

Step 2

Check that point P lies on the circle

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Answer

Substitute the coordinates of point P, (2.1, 2.8), into the equation of the circle: 2.12+2.82=12.252.1^2 + 2.8^2 = 12.25 Calculating gives: 4.41+7.84=12.25,4.41 + 7.84 = 12.25, which confirms that point P lies on the circle.

Step 3

Find the slope of the radius at point P

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Answer

The slope of the radius from the center (0,0) to point P (2.1, 2.8) is given by: ext{slope} = rac{y_2 - y_1}{x_2 - x_1} = rac{2.8 - 0}{2.1 - 0} = rac{2.8}{2.1}.

To simplify, divide both by 0.1: ext{slope} = rac{28}{21} = rac{4}{3}.

Step 4

Determine the slope of the tangent line L

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Answer

The slope of line L, the tangent, is the negative reciprocal of the slope of the radius: ext{slope of L} = - rac{1}{ ext{slope of radius}} = - rac{3}{4}.

Step 5

Use point-slope form to find the equation of L

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Answer

Using point-slope form y−y1=m(x−x1)y - y_1 = m(x - x_1) with point P (2.1, 2.8) and slope - rac{3}{4}: y - 2.8 = - rac{3}{4}(x - 2.1). Distributing gives: y - 2.8 = - rac{3}{4}x + rac{6.3}{4}, which simplifies to: y = - rac{3}{4}x + 1.575 + 2.8. Calculating 2.8+1.575=4.3752.8 + 1.575 = 4.375, leads us to: y = - rac{3}{4}x + 4.375.

Step 6

Convert to standard form ax + by = c

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Answer

To convert to standard form: Multiply through by 4 to eliminate the fraction: 4y=−3x+17.5.4y = -3x + 17.5. Rearranging gives: 3x+4y=17.5.3x + 4y = 17.5. Since a, b, and c must be integers, multiply through by 2 to clear the decimal: 6x+8y=35.6x + 8y = 35. Thus, the equation of line L is: 6x+8y=35.6x + 8y = 35.

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