A circle has centre C(8,12). The point P(5,13) lies on the circle as shown. (a) Find the equation of the tangent at P. The tangent from P meets the y-axis at the point T. (b) (i) State the coordinates of T. (ii) Find the equation of the circle that passes through the points C, P, and T.

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A circle has centre C(8,12) - Scottish Highers Maths - Question 15 - 2019

Question 15

A circle has centre C(8,12). The point P(5,13) lies on the circle as shown. (a) Find the equation of the tangent at P. The tangent from P meets the y-axis at the p... show full transcript

Worked Solution & Example Answer:A circle has centre C(8,12) - Scottish Highers Maths - Question 15 - 2019

Step 1

Find the equation of the tangent at P.

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Answer

To find the equation of the tangent at point P(5,13), we first need to determine the gradient of the radius CP. Given the coordinates of C(8,12) and P(5,13), the gradient of the radius (m) is calculated as follows:

$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{13 - 12}{5 - 8} = \frac{1}{-3} = -\frac{1}{3}$

Since the tangent is perpendicular to the radius at point P, the gradient of the tangent (m_t) will be the negative reciprocal:

$m_t = -\frac{1}{m} = 3$

Now, using the point-slope form of a line equation:

$y - y_1 = m_t(x - x_1)$

Substituting P(5,13):

$y - 13 = 3(x - 5)$

Simplifying gives:

$y = 3x - 15 + 13$ $y = 3x - 2$

Thus, the equation of the tangent at P is:

$y = 3x - 2$

Step 2

State the coordinates of T.

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Answer

The tangent from point P meets the y-axis at point T. Since any point on the y-axis has an x-coordinate of 0, we can find the y-coordinate by substituting x = 0 into the tangent equation:

$0 = 3(0) - 2 \Rightarrow y = -2$

Thus, the coordinates of point T are T(0, -2).

Step 3

Find the equation of the circle that passes through the points C, P, and T.

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Answer

To find the equation of the circle that passes through points C(8,12), P(5,13), and T(0,-2), we will first find the midpoint of line CT and the radius.

Midpoint M of line CT: $M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{8 + 0}{2}, \frac{12 - 2}{2} \right) = (4, 5)$
Radius (r) from C to T: $r = CT = \sqrt{(8 - 0)^2 + (12 - (-2))^2} = \sqrt{64 + 196} = \sqrt{260} = \sqrt{65}$
Thus, the equation of the circle in standard form is:

$(x - 4)^2 + (y - 5)^2 = 65$

Therefore, the equation of the circle that passes through points C, P, and T is:

$(x - 4)^2 + (y - 5)^2 = 65$

A circle has centre C(8,12) - Scottish Highers Maths - Question 15 - 2019

Worked Solution & Example Answer:A circle has centre C(8,12) - Scottish Highers Maths - Question 15 - 2019

Find the equation of the tangent at P.

State the coordinates of T.

Find the equation of the circle that passes through the points C, P, and T.

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