Answer:
Part (a)
Equation of a circle
[tex](x-a)^2+(y-b)^2=r^2[/tex]
where:
- (a, b) is the center
- r is the radius
Given equation: [tex]x^2+y^2=6.25[/tex]
Comparing the given equation with the general equation of a circle, the given equation is a circle with:
- center = (0, 0)
- radius = [tex]\sqrt{6.25}=2.5[/tex]
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.
Part (b)
Given equation: [tex]x+y=1.5[/tex]
Rearrange the given equation to make y the subject: [tex]y=-x+1.5[/tex]
Find two points on the line:
[tex]x=-2 \implies -(-2)+1.5=3.5\implies (-2,3.5)[/tex]
[tex]x=2 \implies -(2)+1.5\implies-0.5\implies (2,-0.5)[/tex]
Plot the found points and draw a straight line through them.
The points of intersection of the circle and the straight line are the solutions to the equation.
To solve this algebraically, substitute [tex]y=-x+1.5[/tex] into the equation of the circle to create a quadratic:
[tex]\implies x^2+(-x+1.5)^2=6.25[/tex]
[tex]\implies x^2+x^2-3x+2.25=6.25[/tex]
[tex]\implies 2x^2-3x-4=0[/tex]
Now use the quadratic formula to solve for x:
[tex]x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0[/tex]
[tex]\implies x=\dfrac{-(-3) \pm \sqrt{(-3)^2-4(2)(-4)}}{2(2)}[/tex]
[tex]\implies x=\dfrac{3 \pm \sqrt{41}}{4}[/tex]
To find the coordinates of the points of intersection, substitute the found values of x into [tex]y=-x+1.5[/tex]
[tex]\implies y=-\left(\dfrac{3 + \sqrt{41}}{4}\right)+1.5=\dfrac{3-\sqrt{41}}{4}[/tex]
[tex]\implies y=-\left(\dfrac{3 - \sqrt{41}}{4}\right)+1.5=\dfrac{3+\sqrt{41}}{4}[/tex]
Therefore, the two points of intersection are:
[tex]\left(\dfrac{3 + \sqrt{41}}{4},\dfrac{3-\sqrt{41}}{4}\right) \textsf{ and }\left(\dfrac{3 - \sqrt{41}}{4},\dfrac{3+\sqrt{41}}{4}\right)[/tex]
Or as decimals to 2 d.p.:
(2.35, -0.85) and (-0.85, 2.35)
