True or False? A radius which passes through a point P on a circle, is perpendicular to the tangent to the circle which passes through point P .

True. A radius which passes through a point P on a circle, is perpendicular to the tangent to the circle which passes through point P .

If the gradient of the radius passing through point P on a circle is known, how can you find the gradient of the tangent to the circle at point P ?

If the gradient of the radius passing through point P on a circle is known, the gradient of the tangent to the circle at point P will be its negative reciprocal . This is because a radius and a tangent which meet at a point, will meet at right angles .

GCSEMathsEdexcelHigherFlashcards2. AlgebraEquation of a Circle

Equation of a Circle (Edexcel GCSE Maths)

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FrontEquation of a Circle
Write down the equation of a circle, centred at the origin, with a radius of $r$ .

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Write down the equation of a circle, centred at the origin, with a radius of $r$ .
The equation of a circle centred at the origin with a radius of $r$ is:
$x^{2} + y^{2} = r^{2}$
How do you find the radius of the circle given by $x^{2} + y^{2} = 81$ ?
To find the radius of the circle given by $x^{2} + y^{2} = 81$ you need to square root 81.
This is because $r^{2} = 81$ from the equation $x^{2} + y^{2} = r^{2}$ .
This gives a radius of 9.
You do not need $\pm 9$ , as a radius is always positive.
Write down the equation of the circle shown.
The equation is $x^{2} + y^{2} = 25$ as the radius is 5 so $r = 5$ in $x^{2} + y^{2} = r^{2}$ .
How would you find the diameter of the circle $x^{2} + y^{2} = 100$ ?
To find the diameter of the circle $x^{2} + y^{2} = 100$ , first compare it to the equation of a circle given by $x^{2} + y^{2} = r^{2}$ .
You can see that the radius, $r$ , is 10, because $r^{2} = 100$ .
The diameter is double the radius. This means the diameter is 20.
True or False?
The radius of the circle $x^{2} + y^{2} = 8$ is $\sqrt{8}$ .
True.
The radius of the circle $x^{2} + y^{2} = 8$ is $\sqrt{8}$ .
It is possible for a radius to be left as a square root.
You can also use surd methods to simplify them, such as $\sqrt{8} = \sqrt{4 \times 2} = \sqrt{4} \times \sqrt{2} = 2 \sqrt{2}$ .
True or False?
The point $(8, 6)$ lies on the circumference of the circle $x^{2} + y^{2} = 100$ .
True.
The point $(8, 6)$ lies on the circumference of the circle $x^{2} + y^{2} = 100$ .
You can show this by substituting in $x = 8$ and $y = 6$ into the left-hand side.
This gives $8^{2} + 6^{2} = 64 + 36 = 100$ .
This matches the right-hand side, so the equation is satisfied (the point lies on the circle).
How can you find out if the point $(4, 3)$ lies inside or outside the circle $x^{2} + y^{2} = 36$ ?
To find out if the point $(4, 3)$ lies inside or outside the circle $x^{2} + y^{2} = 36$ , first substitute $x = 4$ and $y = 3$ into the left-hand side of the equation.
This gives $4^{2} + 3^{2} = 16 + 9 = 25$ .
Compare this number to the right-hand side of the equation. If it is less than the right-hand side, the point lies inside the circle.
$25 < 36$ so the point lies inside the circle.
True or False?
The $y$ -intercepts of the circle $x^{2} + y^{2} - 2 = 0$ are $\pm 2$ .
False.
The equation rearranges to $x^{2} + y^{2} = 2$ .
This means the radius is $\sqrt{2}$ (because $x^{2} + y^{2} = r^{2}$ and $r^{2} = 2$ ).
The $y$ -intercepts of a circle centred at the origin are $\pm r$ (plus or minus the radius).
Therefore the $y$ -intercepts are $\pm \sqrt{2}$ .
True or False?
The equation $x^{2} + y^{2} + 1 = 0$ represents a circle.
False.
The equation rearranges to $x^{2} + y^{2} = - 1$ , which has a negative right-hand side.
The equation of a circle is $x^{2} + y^{2} = r^{2}$ , so the right-hand side must always be positive.
Explain why the line $x = 12$ never intersects the circle $x^{2} + y^{2} = 100$ .
The circle $x^{2} + y^{2} = 100$ is centred at the origin with a radius of 10.
The line $x = 12$ is a vertical line through 12 on the $x$ -axis.
This will never intersect the circle, as the circle lies between $\pm 10$ on the $x$ -axis.
True or False?
A radius which passes through a point P on a circle, is perpendicular to the tangent to the circle which passes through point P.
True.
A radius which passes through a point P on a circle, is perpendicular to the tangent to the circle which passes through point P.
If the gradient of the radius passing through point P on a circle is known, how can you find the gradient of the tangent to the circle at point P?
If the gradient of the radius passing through point P on a circle is known, the gradient of the tangent to the circle at point P will be its negative reciprocal.
This is because a radius and a tangent which meet at a point, will meet at right angles.
Given a point $(a, b)$ on the circumference of a circle with centre $(0, 0)$ , outline how you would find the equation of the tangent passing through $(a, b)$ .
Given a point $(a, b)$ on the circumference of a circle with centre $(0, 0)$ , outline how to find the equation of the tangent passing through $(a, b)$ :
- Find the gradient of the radius: $\frac{b - 0}{a - 0} = \frac{b}{a}$ .
- Use the fact that the tangent is perpendicular to this, so has gradient $- \frac{a}{b}$ .
- The equation of the tangent will be in the form $y = m x + c$ .
- Substitute the gradient $- \frac{a}{b}$ , and the point $(a, b)$ into $y = m x + c$ to find $c$

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