Answer:
<h3><u>Part (a)</u></h3>
<u />
<u>Equation of a circle</u>

where:
- (a, b) is the center
- r is the radius
Given equation: 
Comparing the given equation with the general equation of a circle, the given equation is a <u>circle</u> with:
- center = (0, 0)
- radius =

To draw the circle, place the point of a compass on the origin. Make the width of the compass 2.5 units, then draw a circle about the origin.
<h3><u>Part (b)</u></h3>
Given equation: 
Rearrange the given equation to make y the subject: 
Find two points on the line:


Plot the found points and draw a straight line through them.
The <u>points of intersection</u> of the circle and the straight line are the solutions to the equation.
To solve this algebraically, substitute
into the equation of the circle to create a quadratic:



Now use the quadratic formula to solve for x:



To find the coordinates of the points of intersection, substitute the found values of x into 


Therefore, the two points of intersection are:

Or as decimals to 2 d.p.:
(2.35, -0.85) and (-0.85, 2.35)
Answers:
1.) -x - 5
2.) 11x^2 + 5x
3.) 13x + 2y +3y^2
4.) -12x - 2
5.) -24
Sorry, it won’t somehow let me attach a picture of my work on paper :(
But hopefully this helps! :)
Left side: 63, 72, 16 right side: 36, 4, 35
264 because you can do 8 divided by 6 and get 1.3 repeating, than you would do 198 times 1.333333333333 and you get 264