Math Nerds Only
I need help for some grade 8 work:
I dont even understand the question!:
"Find the equation of the tangent at (0,2) to the circle with the following equation,
(x+2)^2 +(y+1)^2=13 "
I'm not looking for the answer but how to do it and approach it.
April 25, 2015
6 Comments • Newest first
[quote=ianzbored]OK GUYS NEVER MIND:
best way to learn is do it urself instead of asking ^.^
and yes i finally got it xp[/quote]
Great
OK GUYS NEVER MIND:
best way to learn is do it urself instead of asking ^.^
and yes i finally got it xp
[quote=ianzbored]Hi. Thanks for your explination.
I sort of see where you are going but i lost you at how to find the center of circle?
I haven't done this math in a while so im guessing you use the point (0,2) that lies on the circle? Then you solve for the centre of the circle. Find equation of that line. Then find the perpindicular of the line/radius which would be the tangent? @.@[/quote]
To get a better understanding, you should graph it first.
From memory the numbers that are in the brackets with x or y tell you how the center point is moved. (x+2) for example moves the center point 2 to the left of the origin.
So for the equation you gave, the center point should be (-2,-1) [url=https://www.wolframalpha.com/input/?i=%28x%2B2%29%5E2+%2B%28y%2B1%29%5E2%3D13] Graphical View [/url]
From there follow the steps of the first poster and you should get the tangent lines equation.
eh you certainly forgot/ don't know the equation of the circle (not intending to offend you or anything, it's just a way that will make you never forget it )
(x-a)^2 + (y-b)^2 = r^2 where (a,b) are the coordinates of the center of the circle and "r" is the radius and yeah (0,2) lies on the circle so we are practically finding the equation of the radius at (0,2) and then finding the equation of the line perpendicular to it, in reality you only need the gradient of the radius at (0,2) to find the equation of the perpendicular line, but well it's always more fun to find the whole equation and actually this method would work even if it was not a circle so I find it better to continue the job till the end.
[quote=RoyaIBIue]First write down the center of the circle (it's good to draw a sketch), from there join the center to the point (0,2), find the equation of the straight line you just drew, then find the gradient of the line perpendicular to that (-1/gradient of first line) finally use y=mx +c where m is the gradient you just found (of the perpendicular line that is) and put y as 2 and x as 0 then solve to find c and then close your book/notebook
Oh and since you didn't get the question, perhaps you first need to get to know what a tangent is... a tangent is a line that touches the circle (or curve) at only one single point and in this case your point is (0,2)[/quote]
Hi. Thanks for your explination.
I sort of see where you are going but i lost you at how to find the center of circle?
I haven't done this math in a while so im guessing you use the point (0,2) that lies on the circle? Then you solve for the centre of the circle. Find equation of that line. Then find the perpindicular of the line/radius which would be the tangent? @.@
First write down the center of the circle (it's good to draw a sketch), from there join the center to the point (0,2), find the equation of the straight line you just drew, then find the gradient of the line perpendicular to that (-1/gradient of first line) finally use y=mx +c where m is the gradient you just found (of the perpendicular line that is) and put y as 2 and x as 0 then solve to find c and then close your book/notebook
Oh and since you didn't get the question, perhaps you first need to get to know what a tangent is... a tangent is a line that touches the circle (or curve) at only one single point and in this case your point is (0,2)