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GoldRush18
The circle C has equation (x-3)^2+(y-4)^2=25. i. State the radius and the coordinates of the centre of C. ii. Find the equation of the tangent at the point (6,8) on C. iii. Calculate the coordinates of the points of intersection of C with the straight line y=2x+3.
where do you have problem??
let y=mx+c be equation of your tangent |dw:1336583832242:dw|
find the value of m2 put the all the values m2, (6,8) in y = mx + c ... and find the value of c. then you have tangent!!
ok what about part 3
\[ (x - 3)^2 + ((2x+3) - 4)^2 = 25 \] Find the value of x, and put the value of x in equation of circle ... and find corresponding y
ok im still on part 2 when it comes to graph and stuff i just cant get it thanks though
you know how to find slope ... ?? from two points??
is that the same thing as the gradient
yes ... \[ m = \frac{y_2 - y_1}{x_2 - x_1}\]
allright you know when two lines are perpendicular, if m1 and m2 be their gradient or slope then \( m_1 * m_2 = -1\) ??
no i didnt know that so what are the values i should include for this part
m1 is just you calculated ... slope(gradient) of that radius!! m2 .. you have to calculate!! using the relation above.
http://www.khanacademy.org/math/algebra/linear-equations-and-inequalitie/v/perpendicular-lines?playlist=Algebra+I+Worked+Examples http://www.mathopenref.com/coordperpendicular.html I am not sure if it will help!!
No .. you should get m1*m2 = -1 (m1 = 4/3) m2 = -3/4
how do i get C which is the y intercept?
you know one point ... so it satisfies that equation m = -3/4 (x,y) = (6,8) y = mx+c, c=??
Oo .. put the values of x=6 and y=8 since (x,y) is solution and we already know this!!
and im stuck once more
put y = 2x+3 on (x-3)^2+(y-4)^2=25 and find the value of x