can you please give me an easy way of solving this, I have 5 more questions similar to this to do but I have no clue how to even start!

Guest Apr 19, 2019

#1**0 **

I'll get ya started....

The equation of a circle in the form (x-h)^2 + (y-k)^2 = r^2 will give you the center h, k

Get the equation to that form

x^2 + 3x + y^2 - 6y = 0 'complete the square' for x and y

(x+ 1.5)^2 + (y -3)^2 = 2.25 + 9

center = -1.5, 3 radius = sqrt 11.25

Can you keep going from here?

ElectricPavlov Apr 19, 2019

#2**0 **

Yes I can thank you, but can you also teach me how to find the equation of the normal please?

Guest Apr 19, 2019

#3**0 **

You know the center -1.5,3 and a point 0,6 work out the equation of the line that connects the two points (remember y = mx+b ?) (EDIT: just work out the slope (m) between the two points ( y2-y1)/(x2-x1)

then m = slope of that line you just calculated....the slope of the normal (perpindicular) line will be - 1/m ....then work out the equation of this line given the point 0,6..... Can you do it?

ElectricPavlov
Apr 19, 2019