Circles & Coordinate Geometry Revision Session AQA Core 1 Maths Wednesday, March 11 th

Starter With a partner, list the following formulae: equation of a line gradient midpoint distance equation of a circle (describe how to find the centre & radius from the equation)

Solutions Equation of a line: or GradientMidpointDistance Equation of a circle: Where the centre is at point and the radius is

Example 1 – The basics of straight lines Given the points A (-3, 7) and B (5, 1), find: (a)the midpoint between points A and B (b)the distance between points A and B (c)the gradient of the line that passes through AB (d)an equation of the line that passes through point A and B (e)the gradient of the line parallel to AB (f)the gradient of the line perpendicular to AB

Solutions (a) midpoint: Given the points A (-3, 7) and B (5, 1), find: (b) distance: (c) gradient: (d) equation of line AB: (e) gradient of line parallel to AB: Note: If two lines with gradients and are parallel, then (f) gradient of line perpendicular to AB: Note: If two lines with gradients and are perpendicular then

Exam Question – June 2013

If point lies on the line Solutions If a point lies on a line, then it should satisfy its equation. Meaning, we can substitute in the given point to find p. then

Solutions Rearrange the given equation into the form m represents the gradient in this form.

Solutions If AC is perpendicular to AB, then Substitute the point A and C and the gradient of AC into the gradient formula in order to find the y-coordinate, k.

Solutions If two lines intersect, then we can find their solution (where they intersect) by solving for x and y simultaneously! Original New To eliminate x, multiply the top by 2 and the bottom by - 3, then add vertically. Substitute into either above equations to find the x coordinate. Therefore, point D lies at

You try! – timed (10 mins) Exam Question – January 2013

Swap solutions & peer assess

Label the Parts of a Circle key terms: centre circumference radius diameter chord normal line tangent line major arc minor arc

Example 2 – The basics of circles A circle with centre C has equation (a)Express this equation in the form where a, b, and k are integers. (b)Find the centre C. (c)Find the radius of the circle (d)The point D has coordinates (-2, -1). Verify that point D lies on the circle. (e)Sketch the circle.

Solutions (a) Start with the given equation and use “completing the square” technique to rearrange into the form

Solutions If the circle’s equation is (b)Then centre lies at (c)and the radius is (d)To show that point D lies on the circle, simply substitute the coordinates into the original equation and show that it does, in fact, equal zero.

Solutions If the circle’s equation is then the sketch looks like

Exam Question – June 2013 You try! – timed (10 mins)

Solutions (a) Start with the given equation and use “completing the square” technique to rearrange into the form

Solutions (b)(i) then the centre lies at If (ii) the radius is

Solutions (c)(i) Sketch of To get full marks, your circle must be in the correct position cut the negative y-axis twice touch the x-axis at x=5 5 must be marked on x-axis OR centre must be marked (either ok) (c)(ii) The coordinates of the point that lies on the circle and is the furthest away from the x-axis would lie here. This is point (5, -14) because the centre is at (5, -7) and we know the radius is 7 units long.

Thus, we need to describe the transformation which maps the circle with equation onto the circle C. Solutions (d) Solution to part (a) is Let’s take a look at the sketches of these two circles…. Be careful with the wording of this type question! We need write a vector that would map the green circle “ONTO” to the black circle. Translated 6 units right 7 units down

With the remaining time, try the past exam questions given in your hand out & ask questions when needed.