Spatial Sense

Experiential Learning, Collaborative Learning, Talk for Learning and building on what others say
Learning Experience: Learners in collaborative groups deduce the equation of a circle given the endpoints of a diameter.
Activity 1: Learners in mixed-ability groups brainstorm and derive various ways to find the equation of the circle given the endpoints of its diameter, and share their findings with the whole class.
- Learners recollect that a diameter subtends an angle of 90° at the circumference of a circle.

If the point on the diameter is 𝐴 and 𝐵, and the perpendicular point is 𝑃, then 𝐴 is perpendicular to 𝐵.
- Learners recollect that the product of the gradient of 𝐴 and 𝐵 is −1.
- Learners brainstorm and establish that simplifying the product of the gradient of 𝐴 𝑎 𝐵 = −1 will result in the formula for finding the equation of a line. That is, If 𝐴(𝑥 ! , 𝑦 ! ), 𝐵(𝑥 ) , 𝑦 ) ) and 𝑃(𝑥, 𝑦), then product of the gradient of 𝐴 𝑎 𝐵 = −1 is given as 𝑦 − 𝑦 ! 𝑦 − 𝑦 ) ∙ = −1 𝑥 − 𝑥 ! 𝑥 − 𝑥 ) (𝑦 − 𝑦 ! )(𝑦 − 𝑦 ) ) + (𝑥 − 𝑥 ! )(𝑥 − 𝑥 ) ) = 0
- Learners in groups extend their knowledge and understanding to find the centre and the radius.
Example 1: Find the equation of a circle through the ends (5,7) and (1,3) of its diameter and find the centre and radius.
Solution: Equation of the circle: (𝑥 − 5)(𝑥 − 1) + (𝑦 − 7)(𝑦 − 3) = 0 𝑥 + 𝑦 − 6𝑥 − 10𝑦 + 26 = 0 ) )
Centre: 2𝑔 = −6 and 2𝑓 = −10, Therefore the centre of the circle is (3,5)
Radius: 𝑟 ) = 3 ) + 5 ) − 26 𝑟 = √8
- Learners apply the concept of midpoint to find the radius of the circle and the equation of the circle using the idea of distance between two points.
Equation of a Circle given three points.
Activity 2: Learners brainstorm finding the equation of a circle given three points using the general equation of the circle or the equation of a circle in standard form.
- Learners discover that, to find the equation of a circle given three points, they have to substitute the three given coordinates in the general equation of a circle and solve the resulting equations simultaneously.
Example: Find the equation of the circle passing through the points; (1,3), (−1,5)and (−1,1).
Solution: General equation 𝑥 ) + 𝑦 ) + 2𝑔 + 2𝑓 + 𝑐 = 0
Substituting the coordinates will result into 10 + 2𝑔 + 6𝑦 + 𝑐 = 0 26 − 2𝑔 + 10𝑦 + 𝑐 = 0 2 − 2𝑔 + 2𝑦 + 𝑐 = 0 Solving it simultaneously, 𝑦 = −3, 𝑔 = 1, 𝑐 = 6 Therefore, the general equation of the circle is 𝑥 ) + 𝑦 ) + 2𝑥 − 6𝑦 + 6 = 0
- Learners in other groups discover that points can be substituted in the standard form of a general equation, and since the radius of the circle is equal, equation 1 can be equated to equation 2, and equation 2 can be equated to equation 3 to obtain two equations that can be solved simultaneously.
Activity 3: Learners in their groups create practical problems, and other groups solve and present their solutions to the class.
Activity 4: Research Work: In groups, learners research and find alternative ways to find the equation of the circle and present the findings in class.
Activity 5: Learners in their collaborative groups recollect what a tangent is and the theorem between the tangent and the radius of a circle.
- Learners recollect that a tangent to a circle is a straight line that touches the circumference of a circle at one point only.
- Learners in their groups establish that a tangent is always perpendicular to the radius at the point of contact (point of tangency). 

- Learners establish that a point lies on the circle if the point (𝑥, 𝑦) satisfies the equation of a circle.
Activity 6: Learners in their collaborative groups discuss and share ideas on how to find the equation of a tangent to a given circle.
- Learners in their collaborative groups share ideas on how to solve a practical question.
Example 1: Find the equation of the tangent through (3,4) and on the circle 𝑥 ) + 𝑦 ) = 25
Solution
- Learners verify that the point lies on the circle. 3 ) + 4 ) = 25
Therefore, (3,4) lies on the circle.
- From the equation, learners deduce that the circle has its centre at the origin, and the radius is 5. Coordinates to be used to find the slope of the radius of the circle are (0,0) and (3,4) ,*( , The slope of the radius = "*( = " " Since the tangent is perpendicular to the radius, the radius of the tangent is − , The equation of the tangent is given as 𝑦 = 𝑚 + 𝑐, (3,4) " ). Therefore, the equation of the tangent is 𝑦 = − , 𝑥 + ,
Activity 7: Learners in their collaborative groups discuss and brainstorm to find the length of a tangent to a given circle from an external point.
- Group leaders lead their members to sketch a diagram to illustrate the length of a tangent to a given circle from an extended point.

- Learner groups brainstorm and apply their knowledge of Pythagoras theorem and find the distance between two points to find the length of the tangent.
Example 1: Find the length of the tangent from the point (2,8) to the circle 𝑥 ) + 𝑦 ) + 4𝑥 − 10𝑦 + 20 = 0.
Solution:
- Learners collaborate in their groups to find the centre and radius of the circle as (−2,5) and 3, respectively.
- Learners use (−2,5) and (2,8) to find the length of the hypotenuse as 5.
- Learners apply Pythagoras theorem to find the length of the tangent as 4.
Activity 8: Learners in a collaborative group discuss the equation of a normal to a given circle
- Through interactions in learners' collaborative groups, learners discover that a normal is a line perpendicular to the tangent.

- Learners establish that since the normal and tangent lines are perpendicular to each other, the products of their gradient is −1 and the equation of the normal is of the form 𝑦 = 𝑚 + 𝑐.
- Learners solve practical examples to consolidate the concept.
Example 1: Find the equation of the normal to the circle 𝑥 ) + 𝑦 ) = 5.
Solution:
- Through interaction in their collaborative groups, learners establish that the circle has its origin at the centre; therefore, the coordinates at the centre is (0,0).
- )*( ! The gradient of the normal/radius is /*( = " .
- To find 𝑐 𝑦 = 𝑚 + 𝑐 1 2 = (6) + 𝑐 3 𝑐 = 0 ! Therefore, the equation of the circle is 𝑦 = 𝑥 "
- Learners in their groups create questions for other groups to solve and present their solutions to the class.
Teaching and Learning Resources:
- Worksheets
- Scientific Calculator
- Technological tools, apps, etc.
- SHS Additional Mathematics Curriculum
Assessment (2.2.1.AS.3). The document marks these depth-of-knowledge levels for this indicator: Level 2 Skills of conceptual understanding.