True or False? A straight line can meet a circle at exactly one point.

True. That happens when the line is a tangent , touching the circle without crossing it. Otherwise a line meets the circle twice or misses it altogether; three points is impossible.

Define a tangent to a circle.

A tangent is a line that touches a circle at a single point without cutting across it. Because it only touches, a tangent meets the circle exactly once rather than twice.

How is a tangent related to the radius where they meet?

They are perpendicular , so they meet at a right angle. That is what lets you get the tangent's gradient from the radius's gradient.

True or False? A tangent to a circle passes through the centre.

False. A tangent touches the outside of the circle, while a line through the centre cuts right across it and meets it twice. The radius drawn to the point of contact does reach the centre, but the tangent itself does not.

What is the first thing to work out when finding a tangent's equation?

The gradient of the radius joining the centre to the point of contact. The tangent's gradient is then the negative reciprocal of that, because the two are perpendicular.

GCSEFurther MathsAQAFlashcards3. Coordinate GeometryEquation of a Circle

Equation of a Circle (AQA GCSE Further Maths): Flashcards

Exam code: 8365

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FrontEquation of a Circle
Complete the equation of a circle with centre $(a, b)$ and radius $r$ :
${(x - a)}^{2} +______=______$

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Cards in this collection (14)

Complete the equation of a circle with centre $(a, b)$ and radius $r$ :
${(x - a)}^{2} +______=______$
The completed equation is:
${(x - a)}^{2} + {(y - b)}^{2} = r^{2}$
Note that the right-hand side is $r^{2}$ rather than $r$ .
Find the centre and radius of ${(x + 4)}^{2} + {(y - 7)}^{2} = 25$ .
The centre is $(- 4, 7)$ , because the numbers in the brackets have the opposite sign to the coordinates.
The radius is $\sqrt{25} = 5$ , since the right-hand side is $r^{2}$ .
True or False?
The circle ${(x + 12)}^{2} + {(y - 9)}^{2} = 73$ has centre $(12, - 9)$ .
False.
The centre is $(- 12, 9)$ : comparing with ${(x - a)}^{2}$ , a bracket reading $x + 12$ means $a = - 12$ .
Its radius is $\sqrt{73}$ , which shows a radius need not be a whole number.
Write the equation of the circle with centre $(5, - 2)$ and radius 7.
Substituting into ${(x - a)}^{2} + {(y - b)}^{2} = r^{2}$ gives ${(x - 5)}^{2} + {(y - (- 2))}^{2} = 7^{2}$ .
Tidying that up gives ${(x - 5)}^{2} + {(y + 2)}^{2} = 49$ .
How do you find where a line meets a circle?
Substitute the line's $y = m x + c$ into the circle's equation, which leaves a quadratic in $x$ .
Solve that for $x$ , then put each value back into the line to get the matching $y$ .
True or False?
A straight line can meet a circle at exactly one point.
True.
That happens when the line is a tangent, touching the circle without crossing it.
Otherwise a line meets the circle twice or misses it altogether; three points is impossible.
Why put the $x$ -values back into the line rather than the circle?
Because the line is linear, so each $x$ gives exactly one $y$ with no square roots involved.
Using the circle would mean solving a quadratic again, and then working out which $y$ pairs with which $x$ .
A circle has centre $(- 3, 2)$ and radius 4. Find where it meets the line $y = x + 1$ .
Substituting gives ${(x + 3)}^{2} + {(x - 1)}^{2} = 16$ , which simplifies to $x^{2} + 2 x - 3 = 0$ .
That factorises to give $x = - 3$ or $x = 1$ , so the points are $(- 3, - 2)$ and $(1, 2)$ .
Define a tangent to a circle.
A tangent is a line that touches a circle at a single point without cutting across it.
Because it only touches, a tangent meets the circle exactly once rather than twice.
How is a tangent related to the radius where they meet?
They are perpendicular, so they meet at a right angle.
That is what lets you get the tangent's gradient from the radius's gradient.
True or False?
A tangent to a circle passes through the centre.
False.
A tangent touches the outside of the circle, while a line through the centre cuts right across it and meets it twice.
The radius drawn to the point of contact does reach the centre, but the tangent itself does not.
What is the first thing to work out when finding a tangent's equation?
The gradient of the radius joining the centre to the point of contact.
The tangent's gradient is then the negative reciprocal of that, because the two are perpendicular.
Complete the gradient of the tangent, where the radius joins $(x_{1}, y_{1})$ to $(x_{2}, y_{2})$ :
$m = - \frac{x_{2} - x_{1}}{______}$
The completed formula is:
$m = - \frac{x_{2} - x_{1}}{y_{2} - y_{1}}$
This is the negative reciprocal of the radius's gradient, written out in a single step.
Once you have the tangent's gradient, how do you write its equation?
Use the point of contact, which lies on the tangent, together with that gradient.
Substituting into $y - y_{1} = m (x - x_{1})$ , or into $y = m x + c$ to find $c$ , both work.

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