Substitute 𝑦 = 1 – 𝑥 into the equation of the circle and solve for 𝑥: Substitute 𝑥 = 5 and 𝑦 = 8 into this equation: Work out the equation of the tangent, giving your answer in the form 𝑎𝑦 + 𝑏𝑥 + 𝑐 = 0. Work out the point at which this tangent meets the circumference of the circle.ĥ. The circle whose equation is (𝑥 – 1) 2 + (𝑦 + 4) 2 = 8 has a tangent with equation 𝑦 + 𝑥 = 1. The circumference of a circle with centre (-1, 2) passes through the point with coordinates (5, 8). Where necessary, give your answer as a surd in its simplest form.ģ. For each circle, work out the radius and the coordinates of the centre. Write down the coordinates of the centre and the radius of each circle.Ģ. The product of the gradients of two perpendicular lines is -1 or, alternatively, you can say the gradient of the tangent is the negative reciprocal of the gradient of the radius.īy substituting 𝑚 𝑡 and (1, 5) into the formula (𝑦 – 𝑦 1) = 𝑚(𝑥 – 𝑥 1), or by some other means, find the equation of the tangent:ġ.

You can then use this to work out the gradient of the tangent, 𝑚 𝑡. The tangent and radius of a circle meet at 90° so begin by working out the gradient of the radius between (3, 2) and (1, 5), 𝑚 𝑟. Work out the equation of this line, giving your answer in the form 𝑎𝑦 + 𝑏𝑥 + 𝑐 = 0. (Note: this is sometimes referred to as “The angle in a semicircle is 90°.”)Ī circle with centre (3, 2) has a tangent at the point with coordinates (1, 5). The angle subtended by the diameter of a semicircle is 90°.The tangent and radius of a circle meet at 90°.Substitute these values back into the equation 𝑦 – 𝑥 = 3 to get the corresponding 𝑦-values (be careful not to mix these up!):įinally, you can use circle theorems to solve problems involving the equation of a circle. Solving this (remember, you can use your calculator) gives you: Then, substitute this into the equation of the circle. To solve the equations simultaneously, begin by rearranging 𝑦 – 𝑥 = 3 to make 𝑦 the subject: You can find the equation of the points where a line or curve meets the circumference of a circle by solving their respective equations simultaneously.įind the exact coordinates of the points of intersection of the line with equation 𝑦 – 𝑥 = 3 and the circle whose equation is (𝑥 + 1) 2 + (𝑦 – 5) 2 = 8. Since the circumference of the circle passes through (3, 0), you can substitute 𝑥 = 3 and 𝑦 = 0 into this equation:Īgain, the equation is 𝑥 2 + (𝑦 – 1) 2 = 10 Since the centre is (0, 1), the equation of the circle can be provisionally written as: Use the distance formula to find its length:Īs you are given the centre coordinates, the equation is: The radius is the line segment joining (3, 0) and (0, 1). Find the equation of this circle, giving your answer in the form (𝑥 – 𝑎) 2 + (𝑦 – 𝑏) 2 = 𝑟 2 where 𝑎, 𝑏 and 𝑟 are constants to be found. The point with coordinates (3, 0) lies on a circle with centre (0, 1). This formula, which is a direct result of Pythagoras’ theorem, tells us that the distance, 𝑑, between two points (𝑥 1, 𝑦 1) and (𝑥 2, 𝑦 2) is given by: It is important to realise that, given the coordinates of the centre of the circle and a point on its circumference, you can find the length of the radius by using the distance formula. Since the radius, 3, is greater than the smallest distance of the centre from the origin, 2, the circle must intersect the 𝑦-axis. The centre is (-2, 4) and its radius is 3 units. Does the circle intersect the axes? Explain your reasoning. Substitute 𝑥 = 1 and 𝑦 = 4 into the equation of the circle found in part a.Ĭ. Show that the point with coordinates (1, 4) lies on the circumference of this circle. The radius squared is 9, so the radius is 3 units.ī. Put this all back into the original equation, collect the constants and rearrange into the equation for a circle: The 𝑥 terms are 𝑥 2 + 4𝑥 so begin by writing this expression in completed square form: Find the coordinates of the centre and the radius of the circle.Ī. If a question does not give the equation of a circle in this form, you will need to complete the square for 𝑥 and 𝑦 to convert it to the general form.Ī circle has equation 𝑥 2 + 4𝑥 + 𝑦 2 – 8𝑦 + 11 = 0.Ī. This gives the equation for a circle with a centre at (𝑎, 𝑏) and a radius of 𝑟: To move the centre 𝑏 up, we replace 𝑦 2 with (𝑦 – 𝑏) 2.To move the centre 𝑎 to the right, we replace 𝑥 2 with (𝑥 – 𝑎) 2.You can generalise this to find the equation of a circle with a centre at any point by applying a transformation to the above graph. Are you looking for info on circles (Maths revision for A level)? A circle, with a centre at the origin and a radius of 𝑟, can be described with the equation:

Circle geometry a level