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Equation of a Circle 2

Equation of a Circle 2

Equation of a Circle 2

Circles can interact with tangents, secants and chords on a graph. We can approach them as simultaneous equations to find the points of intersection.

Circles can interact with straight line functions on a graph. If there are points of intersection, we can find them be solving simultaneous equations.

There are three main ways that a straight line can interact with a circle. These include secants, chords and tangents

Algebraically, we can find whether a line meets a circle at one, two or no points, by solving the equation of a circle and line simultaneously by substitution. This will give us the coordinates of intersection if there are any to be found.

If you haven't done so yet, check out this lesson on solving simultaneous equations. It will help you to understand the next part!

For example, let's find the coordinates where the following circle and the line y=xy=x meet

(x1)2+(y2)2=25(x - 1)^2 +(y-2)^2 = 25


Substitute x for y in the equation of the circle.

(x1)2+(x2)2=25(x - 1)^2 +(x-2)^2 = 25


Expand the brackets on the left hand side



Simplify and subtract 25 from both sides. x22x+1+x24x+4=25x^2-2x+1+x^2-4x+4=25


2 is a common factor of this equation

2x26x20=0  x23x10=0{\cancel{2}}x^2-{\cancel{6}}x-{\cancel{20}}=0 \space\rightarrow \space x^2-3x-10=0


Factorise x23x10=0x^2-3x-10=0


Use the factorisation to find solutions

x=5x=5 and x=2x=-2


Remember that the line is y=xy=x

Therefore, each yy coordinate is the same as the xx


Now we know our coordinates!

The lines meet at (5,5)(5, 5) and (2,2)(-2,-2)

The line y=3y=3 and the following circle cross at which point(s)? x2+y2=25x^2 + y^2 = 25

We may also want to find the equation of a tangent line which meets a circle at a particular point.

For example, the point (2,2)(2, 2) lies on the following circle x2+y2=8x^2 + y^2 = 8


First, find the gradient of the radius

We know that the radius goes through (2,2)(2,2), and will be perpendicular to the tangent at this point.


Identify the centre of the circle

The centre is at (0,0)(0,0). The radius of the circle will also pass through this point.


Find the gradient using the formula

Gradient = 2020=22=1\dfrac{2-0}{2-0}=\dfrac{2}{2}=1


Find the gradient of the tangent

The tangent is perpendicular to the radius. Therefore, their gradients must multiply to make 1-1.


What is the gradient of the tangent, if the radius gradient is 1?


We have the gradient, and a point on the tangent

Now we can substitute these values into y=mx+cy=mx+c to find the equation of the tangent.


Substitute the values into the general equation

m=1m=-1, so at (2,2)(2,2) - 2=1(2)+c2=-1(2)+c


What is c if 2=1(2)+c2=-1(2)+c?



The equation of the tangent line is y=x+4y=-x+4

The gradient of the tangent at the point (4, 4) for the following circle is..

(x1)2+y2=25(x-1)^2 + y^2 = 25

The gradient of the tangent to the following circle is 34-\dfrac{3}{4} at the point (4,4)(4,4). What is the equation of the tangent? (x1)2+y2=25(x-1)^2 + y^2 = 25