can u help
this the qustion is :
A circle passes through the points (1,4)&(5,0) and its centre lies on the line x+y-3=0.Find
(i) the equation of the circle and it parametric equations .
(ii) the centre,diameter and area of the circle .
(iii) the equation of the tangent at (1,4) .
(iv) the equation of the circle concentric with the above circle and passing through the point (7,8).
let A be (1,4) and B be (5,0)
So AB must be a chord of the circle and the centre must be the intersection of the right-bisector of AB and the line x+y - 3 = 0
slope of AB = -4/4 = -1
slope of right-bisector = 1
midpoint of AB = (3,2)
equation of right-bisector: y = x + b
with (3,2) on it
2 = 3 + b ---> b = -1
right-bisector: y = x-1
solve with the second equation:
x+y - 3 = 0
x + (x-1) = 3
2x = 4
x = 2, then y = 3
centre is (2,3)
radius is distance from centre to A
= √((2-1)^2 + (3-4)^2) = √2
i) equation of circle: (x-2)^2 + (y-3)^2 = 2
ii) you do it
iii) slope of radius from A to centre = ...
slope of tangent at A is ..... (the negative reciprocal of above
now you have a point and slope of the tangent....
iv) equation will be
(x-2)^2 + (y-3)^ = r^2
plug in (7,8) to get r^2
Just noticed that you posted the same question twice, and that both Steve and I answered it.
BUT, I made a stupid error in my centre calculation, (that's what you get when you try to do things in your head without writing them down)
I had the centre as (2,3), should have been (2,1) like Steve had
so go with his solution.
Heh heh -- We all do these problems just for fun anyway. Any help to the students is just a welcome side effect. And, these little boo-boos help keep us humble...
Of course, I can help you! Let's break down each part of the question one by one:
(i) To find the equation of the circle, we first need to find the center and radius of the circle. Since the center of the circle lies on the line x+y-3=0, we can find the coordinates of the center by solving this equation along with the equation of the perpendicular bisector of the line segment joining the two given points (1,4) and (5,0).
Let's find the midpoint of the line segment joining the two given points:
Midpoint = ((x1 + x2)/2, (y1 + y2)/2)
Midpoint = ((1 + 5)/2, (4 + 0)/2)
Midpoint = (3, 2)
The slope of the line joining the two points is given by:
m = (y2 - y1)/(x2 - x1)
m = (0 - 4)/(5 - 1)
m = -1
Since the line joining the two points is perpendicular to the line x+y-3=0, the slope of the perpendicular bisector is the negative reciprocal of -1, which is 1.
Using the slope-intercept form of a line, y = mx + b, we can substitute the coordinates of the midpoint (3, 2) and solve for b:
2 = 1(3) + b
2 = 3 + b
b = -1
Therefore, the equation of the perpendicular bisector is y = x - 1.
Now, we can solve the system of equations between the line x+y-3=0 and y = x - 1 to find the coordinates of the center of the circle.
Substituting y = x - 1 in the equation x + y - 3 = 0:
x + (x - 1) - 3 = 0
2x - 4 = 0
x = 2
Substituting x = 2 in the equation y = x - 1:
y = 2 - 1
y = 1
Therefore, the center of the circle is (2, 1).
To find the radius, we can use the distance formula between the center of the circle (2, 1) and one of the given points (1, 4):
radius = sqrt((x2 - x1)^2 + (y2 - y1)^2)
radius = sqrt((2 - 1)^2 + (1 - 4)^2)
radius = sqrt(1 + 9)
radius = sqrt(10)
Now we have the center (2, 1) and the radius sqrt(10). We can use this information to write the equation of the circle in standard form:
(x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center and r is the radius.
Plugging in the values, we get:
(x - 2)^2 + (y - 1)^2 = 10
To find the parametric equations of the circle, we can write x and y in terms of the angle θ:
x = h + r * cos(θ)
y = k + r * sin(θ)
In this case, the parametric equations would be:
x = 2 + sqrt(10) * cos(θ)
y = 1 + sqrt(10) * sin(θ)
(ii) Now, let's find the center, diameter, and area of the circle.
The center of the circle is (2, 1) as we found earlier.
The diameter of a circle is twice the radius. So, the diameter = 2 * sqrt(10).
To find the area of the circle, we can use the formula:
Area = π * r^2, where r is the radius.
Area = π * (sqrt(10))^2
Area = 10π
Therefore, the center is (2, 1), the diameter is 2 * sqrt(10), and the area is 10π.
(iii) To find the equation of the tangent at (1, 4), we need to find the slope of the line perpendicular to the radius passing through (1, 4). The slope of a line perpendicular to another line with slope m is -1/m.
The slope of the radius can be found by taking the negative of the slope of the line joining the center (2, 1) and (1, 4):
m = (4 - 1)/(1 - 2) = 3
Therefore, the slope of the radius is 3. The slope of the tangent at (1, 4) is then -1/3.
Using the point-slope form of a line, y - y1 = m(x - x1), we can substitute the values (x1, y1) = (1, 4) and m = -1/3 to find the equation of the tangent:
y - 4 = (-1/3)(x - 1)
y - 4 = (-1/3)x + 1/3
y = (-1/3)x + 13/3
Therefore, the equation of the tangent at (1, 4) is y = (-1/3)x + 13/3.
(iv) Lastly, to find the equation of the circle concentric with the above circle and passing through the point (7, 8), we can use the same center as the original circle, which is (2, 1).
To find the radius of the new circle, we can use the distance formula between the center (2, 1) and the given point (7, 8):
radius = sqrt((7 - 2)^2 + (8 - 1)^2)
radius = sqrt(25 + 49)
radius = sqrt(74)
Now we have the center (2, 1) and the radius sqrt(74). We can use this information to write the equation of the circle in standard form:
(x - h)^2 + (y - k)^2 = r^2, where (h, k) is the center and r is the radius.
Plugging in the values, we get:
(x - 2)^2 + (y - 1)^2 = 74
This is the equation of the circle concentric with the original circle and passing through the point (7, 8).