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The point (-5,2) lies on the circumference of a circle, centre (0, 0) - OCR - GCSE Maths - Question 19 - 2019 - Paper 6

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The point (-5,2) lies on the circumference of a circle, centre (0, 0). (a) Find the equation of the circle. (b) Work out the gradient of the tangent to the circle ... show full transcript

Worked Solution & Example Answer:The point (-5,2) lies on the circumference of a circle, centre (0, 0) - OCR - GCSE Maths - Question 19 - 2019 - Paper 6

Step 1

Find the equation of the circle.

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Answer

To find the equation of the circle, we will use the standard form of the equation of a circle, which is given by:

x2+y2=r2x^2 + y^2 = r^2

Here,

  • The center of the circle is at (0,0), so the equation simplifies to x2+y2=r2x^2 + y^2 = r^2.
  • We need to find the radius, rr, which is the distance from the center to the point (-5,2).

Using the distance formula:

r=extDistance=extsqrt((x2x1)2+(y2y1)2)r = ext{Distance} = ext{sqrt}((x_2 - x_1)^2 + (y_2 - y_1)^2)

Substituting the values:

r=extsqrt((50)2+(20)2)=extsqrt(25+4)=extsqrt(29)r = ext{sqrt}((-5 - 0)^2 + (2 - 0)^2) = ext{sqrt}(25 + 4) = ext{sqrt}(29)

Therefore, the equation of the circle becomes:

x2+y2=29x^2 + y^2 = 29

Step 2

Work out the gradient of the tangent to the circle at (-5,2).

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Answer

To find the gradient of the tangent at the point (-5,2), we need to first find the gradient of the radius at that point.

The gradient (or slope) of the radius from the center (0,0) to the point (-5,2) is calculated as follows:

m=y2y1x2x1=2050=25=25m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 0}{-5 - 0} = \frac{2}{-5} = -\frac{2}{5}

The tangent to the circle is perpendicular to the radius, so its gradient is the negative reciprocal of the radius gradient:

mtangent=1mradius=125=52m_{tangent} = -\frac{1}{m_{radius}} = -\frac{1}{-\frac{2}{5}} = \frac{5}{2}

Thus, the gradient of the tangent to the circle at (-5,2) is:

52\frac{5}{2}

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