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A circle C with centre at the point (2, -1) passes through the point A at (4, -5) - Edexcel - A-Level Maths Pure - Question 4 - 2015 - Paper 2

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A circle C with centre at the point (2, -1) passes through the point A at (4, -5). (a) Find an equation for the circle C. (b) Find an equation of the tangent to th... show full transcript

Worked Solution & Example Answer:A circle C with centre at the point (2, -1) passes through the point A at (4, -5) - Edexcel - A-Level Maths Pure - Question 4 - 2015 - Paper 2

Step 1

Find an equation for the circle C.

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Answer

To find the equation of a circle given its center and a point on the circle, we use the standard equation:
(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.

  1. Identify the center and point:
    The center (h,k)=(2,−1)(h, k) = (2, -1) and the point A (4,−5)(4, -5) is on the circle.

  2. Calculate the radius:
    The radius is the distance between the center and point A, which can be calculated using the distance formula:
    r=extDistance=extsqrt((x2−x1)2+(y2−y1)2)r = ext{Distance} = ext{sqrt}((x_2 - x_1)^2 + (y_2 - y_1)^2)
    Substituting the coordinates:
    r=extsqrt((4−2)2+(−5+1)2)=extsqrt(22+(−4)2)=extsqrt(4+16)=extsqrt(20)=2extsqrt(5)r = ext{sqrt}((4 - 2)^2 + (-5 + 1)^2) = ext{sqrt}(2^2 + (-4)^2) = ext{sqrt}(4 + 16) = ext{sqrt}(20) = 2 ext{sqrt}(5)

  3. Formulate the equation:
    Substitute hh, kk, and rr into the circle's equation:
    (x−2)2+(y+1)2=(2extsqrt(5))2(x - 2)^2 + (y + 1)^2 = (2 ext{sqrt}(5))^2
    Simplifying gives:
    (x−2)2+(y+1)2=20(x - 2)^2 + (y + 1)^2 = 20
    Thus, the equation of the circle C is:
    (x−2)2+(y+1)2=20(x - 2)^2 + (y + 1)^2 = 20

Step 2

Find an equation of the tangent to the circle C at the point A.

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Answer

To find the equation of the tangent to the circle at point A, we will utilize the following steps:

  1. Find the gradient of the radius:
    The gradient from the center (2,−1)(2, -1) to point A (4,−5)(4, -5) is given by:
    extGradient=y2−y1x2−x1=−5−(−1)4−2=−42=−2ext{Gradient} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-5 - (-1)}{4 - 2} = \frac{-4}{2} = -2

  2. Determine the gradient of the tangent:
    The tangent line is perpendicular to the radius, and the product of their gradients is -1:
    mtangent=−1mradius=−1−2=12m_{tangent} = -\frac{1}{m_{radius}} = -\frac{1}{-2} = \frac{1}{2}

  3. Use point-slope form to find the tangent line equation:
    Using the point-slope formula:
    y−y1=m(x−x1)y - y_1 = m(x - x_1)
    where (x1,y1)=(4,−5)(x_1, y_1) = (4, -5) and m=12m = \frac{1}{2}:
    y+5=12(x−4)y + 5 = \frac{1}{2}(x - 4)
    Multiplying by 2 to eliminate the fraction:
    2(y+5)=x−42(y + 5) = x - 4
    Rearranging gives:
    −x+2y+10+4=0-x + 2y + 10 + 4 = 0
    Therefore, the tangent line can be expressed as:
    x−2y−14=0x - 2y - 14 = 0
    Thus, the equation of the tangent line at point A is:
    x−2y−14=0x - 2y - 14 = 0

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