L is the circle with equation $x^2 + y^2 = 4$
$P\left(\frac{3}{2}, \frac{\sqrt{7}}{2}\right)$ is a point on L - Edexcel - GCSE Maths - Question 23 - 2017 - Paper 2
Question 23
L is the circle with equation $x^2 + y^2 = 4$
$P\left(\frac{3}{2}, \frac{\sqrt{7}}{2}\right)$ is a point on L.
Find an equation of the tangent to L at the point P.
Worked Solution & Example Answer:L is the circle with equation $x^2 + y^2 = 4$
$P\left(\frac{3}{2}, \frac{\sqrt{7}}{2}\right)$ is a point on L - Edexcel - GCSE Maths - Question 23 - 2017 - Paper 2
Step 1
Find the Gradient of the Radius OP
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Answer
To find the gradient of the radius from the origin O to the point P, we first determine the coordinates of P:
P(23,27)
The coordinates of O are (0,0). The gradient (m) can be calculated using the formula:
m=x2−x1y2−y1=23−027−0=3/27/2=37
Step 2
Find the Gradient of the Tangent
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Answer
The gradient of the tangent line at point P is the negative reciprocal of the radius OP's gradient. Thus:
mt=−m1=−73
Step 3
Formulate the Equation of the Tangent Line
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Answer
Using the point-slope form of a line, the equation of the tangent at point P ((\frac{3}{2}, \frac{\sqrt{7}}{2})) can be expressed as:
y−y1=mt(x−x1)
Substituting the known values:
y−27=−73(x−23)
This can be rearranged to find the final equation of the tangent line.
