Question

Please don't answer wrong.​

Answer 1

Part (a)

Equation of a circle

`(x-a)^2+(y-b)^2=r^2`

where:

• (a, b) is the center
• r is the radius

Given equation:
`x^2+y^2=6.25`

Comparing the given equation with the general equation of a circle, the given equation is a circle with:

• center = (0, 0)
• radius =
  `√(6.25)=2.5`

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:
`x+y=1.5`

Rearrange the given equation to make y the subject:
`y=-x+1.5`

Find two points on the line:

`x=-2 \implies -(-2)+1.5=3.5\implies (-2,3.5)`

`x=2 \implies -(2)+1.5\implies-0.5\implies (2,-0.5)`

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
`y=-x+1.5` into the equation of the circle to create a quadratic:

`\implies x^2+(-x+1.5)^2=6.25`

`\implies x^2+x^2-3x+2.25=6.25`

`\implies 2x^2-3x-4=0`

Now use the quadratic formula to solve for x:

`x=(-b \pm √(b^2-4ac) )/(2a)\quad\textsf{when }\:ax^2+bx+c=0`

`\implies x=(-(-3) \pm √((-3)^2-4(2)(-4)))/(2(2))`

`\implies x=(3 \pm √(41))/(4)`

To find the coordinates of the points of intersection, substitute the found values of x into
`y=-x+1.5`

`\implies y=-\left((3 + √(41))/(4)\right)+1.5=(3-√(41))/(4)`

`\implies y=-\left((3 - √(41))/(4)\right)+1.5=(3+√(41))/(4)`

Therefore, the two points of intersection are:

`\left((3 + √(41))/(4),(3-√(41))/(4)\right) \textsf{ and }\left((3 - √(41))/(4),(3+√(41))/(4)\right)`

Or as decimals to 2 d.p.:

(2.35, -0.85) and (-0.85, 2.35)