Photo AI

L is the circle with equation $x^2+y^2=4$ - Edexcel - GCSE Maths - Question 23 - 2017 - Paper 2

Question icon

Question 23

L-is-the-circle-with-equation-$x^2+y^2=4$-Edexcel-GCSE Maths-Question 23-2017-Paper 2.png

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... show full transcript

Worked Solution & Example Answer:L is the circle with equation $x^2+y^2=4$ - Edexcel - GCSE Maths - Question 23 - 2017 - Paper 2

Step 1

Find the Gradient of the Radius at Point P

96%

114 rated

Answer

The radius of the circle at point P(32,72)P\left(\frac{3}{2},\frac{\sqrt{7}}{2}\right) connects the center of the circle (0,0)(0,0) to point PP. The gradient of this radius can be calculated as:

mOP=y2−y1x2−x1=72−032−0=73m_{OP} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\frac{\sqrt{7}}{2} - 0}{\frac{3}{2} - 0} = \frac{\sqrt{7}}{3}

This is the gradient of the radius from the center to point PP.

Step 2

Find the Gradient of the Tangent Line

99%

104 rated

Answer

The tangent line at point PP is perpendicular to the radius OPOP. Therefore, the gradient of the tangent line can be determined using the negative reciprocal of the gradient of the radius:

mtangent=−1mOP=−37m_{tangent} = -\frac{1}{m_{OP}} = -\frac{3}{\sqrt{7}}

Step 3

Equation of the Tangent Line at Point P

96%

101 rated

Answer

Using the point-slope form of a line equation, which is given by:

y−y1=m(x−x1)y - y_1 = m(x - x_1)

Substituting m=−37m = -\frac{3}{\sqrt{7}}, x1=32x_1 = \frac{3}{2}, and y1=72y_1 = \frac{\sqrt{7}}{2} gives:

y−72=−37(x−32)y - \frac{\sqrt{7}}{2} = -\frac{3}{\sqrt{7}} \left(x - \frac{3}{2}\right)

This equation can be rearranged to find the explicit form of the tangent line.

Join the GCSE students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;