cartesian plane...?

Question

Question:
If the two ends points of the diameter of a circle on the Cartesian plane are P(9,-1) and Q(-3,-7) find

a) the coordinates of the centre of the circle
b) the exact length of the radius of the circle

Note: Please take me though how to do this and how you got your answers. Thankyou for your help.

Answer 1

Solution strategy:
Since PQ is the diameter of the circle then the center of the circle will be lying on the middle of PQ. Also the length of the diameter will be double as the radius of the circle

Theory:
The coordinates of the middle point of a line segment AB where A(x1,y1) and B(x2,y2) are the mean values of the respective coordinates of A and B. So the middle pooint C((x1+x2)/2, (y1+y2)/2).

Practice:
For PQ we have C((9-3)/2,(-1-7)/2), C(6/2,-8/2)

C(3,-4) is the center of the circle.

Theory:
The lenght of a line segment AB where A(x1,y1) and B(x2,y2) is the following: dist(A,B)=square_root((x1-x2)^2 + (y1-y2)^2).

Practice:
For PQ we have
dist(P,Q)=square_root((9-(-3))^2+(-1-(-7))^2)
=square_root((9+3)^2+(6)^2)
=square_root(12^2+6^2)
=square_root(144+36)
=square_root(180)
=square_root(4*5*9)
=square_root(2^2*5*3^2)
=square_root(5)*6 exactly or
=13.416407864998738178455042012388... approximately

So the radius is half that size so r=square_root(5)*3 exactly or 6.7082039324993690892275210061938... approximately

Answers:
a)C(3,-4)
b)r=square_root(5)*3 exactly or 6.7 approximately

Answer 2

The diameter of a circle, by definition, is the chord that passes through the circle's center. So the mid point of this must be the circle's center. The point between P and Q is ( (9+(-3))/2, (-1+(-7))/2), or (3, -4).

Now we just have to find the radius. This is going to be half the length of the diameter. Or we can calculate it as the distance between the center point to either end of the diameter. The distance between (3,-4) and (-3,-7) is, using the distance formula: √[ (3-(-3))² + (-4-(-7))²] = √[ 6² + 3² ] = √(45) =3√5.

Notice that you get the same answer if you take the diameter length and cut it in half: √[ (9-(-3))² + (-1-(-7))² ] = √(12² + 6²) = √(180) = √(36*5) = 6√5, so half of this is 3√5.

You have the center point and raidus of the circle, so now you can write the equation if you wanted to:
(x-3)² + (y+4)² = 45

Answer 3

(a) The centre of the circle is the mid-point of PQ. Its co-ordinates are x = (9-3)/2 = 3, y = (-7-1)/2 = -4. Its x co-ordinate is the average of the two x co-ordinates, and its y co-ordinate is the average of the two y co-ordinates.

(b) The radius is half the distance from P to Q. Using Pythagoras' theorem, this is
(1/2)sqrt((9+3)^2 + (-1+7)^2)) = (1/2)sqrt(144+36)
= (1/2)sqrt(180) = (1/2)sqrt(36*5) = 3sqrt5.

Answer 4

the tale is going that Rene Descartes (the "i think of, as a result i'm" guy) observed a fly crawling on the ceiling, and found out that he ought to symbolize 2 factors (say, speed and distance) in a grid. many times the grid has 4 squares, the place the right spectacular is advantageous for the two variables, the decrease spectacular and better left are for circumstances the place one variable is damaging, and the decrease left is the place the two are damaging. maximum instances, once you notice a graph, you're seeing one sq. of the Cartesian plane (additionally time-commemorated via fact the Cartesian co-ordinates); it rather is maximum consumer-friendly via fact for a impressive style of issues in existence--say, my occasion of speed and distance--you do no longer many times could handle damaging numbers. i think of that is called a "plane" via fact it basically has 2 dimensions, rather top and width