Question

Please don't answer wrong.​

Answer 1

Answer:

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 = $\sqrt{6.25}=2.5$

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:

$x=\dfrac{-b \pm \sqrt{b^2-4ac} }{2a}\quad\textsf{when }\:ax^2+bx+c=0$

$\implies x=\dfrac{-(-3) \pm \sqrt{(-3)^2-4(2)(-4)}}{2(2)}$

$\implies x=\dfrac{3 \pm \sqrt{41}}{4}$

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

$\implies y=-\left(\dfrac{3 + \sqrt{41}}{4}\right)+1.5=\dfrac{3-\sqrt{41}}{4}$

$\implies y=-\left(\dfrac{3 - \sqrt{41}}{4}\right)+1.5=\dfrac{3+\sqrt{41}}{4}$

Therefore, the two points of intersection are:

$\left(\dfrac{3 + \sqrt{41}}{4},\dfrac{3-\sqrt{41}}{4}\right) \textsf{ and }\left(\dfrac{3 - \sqrt{41}}{4},\dfrac{3+\sqrt{41}}{4}\right)$

Or as decimals to 2 d.p.:

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