Easy questions (1 Viewer)

[Danger]

I need help with the following:

1. If the distance between (a,3) and (4,2) is square root of 37, find the values of a.
ANSWER: a = -10, 2

2. Find the exact length of the diameter of a circle with centre (-3,4) if the circle passes through the point (7,5)
ANSWER: 2(root 101)

btw, how do you type square root signs, indices etc?

 

[Mattamz]

1. If the distance between (a,3) and (4,2) is square root of 37, find the values of a.

Using the distance formula:
sqrt(37) = sqrt[(a-4)^2+(3-2)^2]
37 = (a-4)^2+(3-2)^2
37 = a^2-8a+16 + 1
0 = a^2 - 8a -20
0 = (a-10)(a+2)
therefore a =10, -2
i think your answer maybe wrong, as its 10, and -2 not -10 and 2.

2. Find the exact length of the diameter of a circle with centre (-3,4) if the circle passes through the point (7,5)

r = sqrt[(7+3)^2+(5-4)^2]
=sqrt[101]
since the diamter is twice the radius, diameter = 2sqrt(101)

 

[Danger]

for question 2, why is it incorrect to have (-3-7)^2 instead of (7+3)^2

 

[sangboi]

for question 2, why is it incorrect to have (-3-7)^2 instead of (7+3)^2

mattamz just subbed numbers into the equation of a circle ie.

(x-h)^2 + (y-k)^2 = r^2

where, (h,k) is the centre of the circle and (x,y) are points that lie on the circle and r is the radius.

since your circle has centre (-3,4) subb that in for your centre (h,k) first and you get.

(x-(-3))^2 + (y-4)^2 = r^2
simplifying...
(x+3)^2 + (y-4)^2 = r^2

now sub in the point (7,5) that the circle goes through the circle as your (x,y).

(7+3)^2 + (5-4)^2 = r^2

now you can find r by calculator work or by hand since its 'exact value'. to get the diameter just multiply r by 2.

and thats how you get your (7+3)^2 i'm not sure what you did to get (-3-7)^2 but the above method is the way to do it.

Hope that helps!

Sang.

 

[tommykins]

It isn't mate, they're both squared, you will have the same result.

 

[Danger]

thank you sang.