Question

Arrange the circles (represented by their equations in general form) in ascending order of their radius lengths.
please help ​

Answer 1

Answer with explanation:

a)

$x^2+y^2-2x+2y-1=0$

It could be expressed as:

$(x-1)^2-1+(y+1)^2-1-1=0\\\\\\i.e.\\\\\\(x-1)^2+(y+1)^2=3\\\\\\i.e.\\\\\\(x-1)^2+(y+1)^2=(\sqrt{3})^2$

Hence, the radius of circle is: √3≈1.732 units

b)

$x^2+y^2-4x+4y-10=0$

It is represented as:

$(x-2)^2-4+(y+2)^2-4-10=0\\\\\\i.e.\\\\\\(x-2)^2+(y+2)^2=18\\\\\\(x-2)^2+(y+2)^2=(3\sqrt{2})^2$

Hence, the radius of circle is: 3√2≈4.242 units

c)

$x^2+y^2-8x-6y-20=0$

on converting to standard form

$(x-4)^2+(y-3)^2=(3\sqrt{5})^2$

Hence, the radius of circle is: 3√5≈6.708 units

d)

$4x^2+4y^2+16x+24y-40=0$

on dividing both side by 4 we obtain:

$x^2+y^2+4x+6y-10=0\\\\\\(x+2)^2+(y+3)^2=(\sqrt{23})^2$

Hence, radius of circle is: √23=4.796 units

e)

$5x^2+5y^2-20x+30y+40=0$

on dividing both side by 5 we obtain:

$x^2+y^2-4x+6y+8=0$

$(x-2)^2+(y+3)^2=(\sqrt{5})^2$

Hence, radius of circle is: √5=2.236 units

f)

$2x^2+2y^2-28x-32y-8=0$

which could also be represented as follows:

$x^2+y^2-14x-16y-4=0\\\\\\(x-7)^2+(y-8)^2=(\sqrt{117})^2$

Hence, the radius of circle is: $\sqrt{117}$≈ 10.817 units

g)

$x^2+y^2+12x-2y-9=0$

It could also be written as:

 $(x+6)^2+(y-1)^2=(\sqrt{46})^2$

Hence, the radius of circle is: $\sqrt{46}$≈ 6.782 units

            The ascending order is:

        a → e → b → d → c → g → f