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# I need help with this please.

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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!

Apr 19, 2019

### 3+0 Answers

#1
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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?

Apr 19, 2019
#2
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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
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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
edited by ElectricPavlov  Apr 19, 2019