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).
The center of the circle also lies on the perpendicular bisector of any chord. We have the chord between the two given points.
The chord has slope (0-4)/(5-1) = -1.
The midpoint of the chord is (3,2).
The equation of the line containing (3,2) and perpendicular to the chord is
(y-2)/(x-3) = 1
or
-x + y = -1
This intersects the line
x + y = 3 at
(2,1)
So, our circle has equation
(x-2)^2 + (y-1)^2 = r^2
plug in either of the given points and we find r^2 = 10. so,
(x-2)^2 + (y-1)^2 = 10
x = 2 + √10 cos t
y = 1 + √10 sin t
the rest should be easy to figure out.
Of course, I can help you with that!
To find the equation of the circle, we need to determine its center and radius.
Step 1: Find the center of the circle.
Since the center of the circle lies on the line x+y-3=0, we need to find the intersection point between this line and the perpendicular bisector of the line joining the given points (1,4) and (5,0).
The midpoint of this line is (3,2), and the slope of the line joining the given points is -1.
The slope of the perpendicular bisector is the negative reciprocal of -1, which is 1.
Using the point-slope form, we have:
y - 2 = 1(x - 3)
y - 2 = x - 3
y = x - 1
To find the intersection point, we need to solve the system of equations:
y = x - 1
x + y - 3 = 0
Substituting y = x - 1 into the second equation:
x + (x - 1) - 3 = 0
2x - 4 = 0
2x = 4
x = 2
Substituting x = 2 into y = x - 1:
y = 2 - 1
y = 1
Therefore, the center of the circle is (2,1).
Step 2: Find the radius of the circle.
We can find the radius by calculating the distance between the center of the circle (2,1) and one of the given points, say (1,4), using the distance formula:
r = √((x2 - x1)^2 + (y2 - y1)^2)
r = √((1 - 2)^2 + (4 - 1)^2)
r = √((-1)^2 + 3^2)
r = √(1 + 9)
r = √10
Now we have the center of the circle (2,1) and the radius √10.
(i) The equation of the circle:
The general equation of a circle is (x - h)^2 + (y - k)^2 = r^2, where (h,k) is the center of the circle.
Substituting the values we found, the equation of the circle is:
(x - 2)^2 + (y - 1)^2 = (√10)^2
(x - 2)^2 + (y - 1)^2 = 10
The parametric equations of the circle can be obtained by expressing x and y in terms of a parameter, usually denoted by θ.
(ii) The center, diameter, and area of the circle:
We already determined the center of the circle, which is (2,1).
The diameter of the circle is twice the radius, so the diameter is 2√10.
To find the area of the circle, we use the formula A = πr^2, where r is the radius. Therefore, the area is 10π.
(iii) The equation of the tangent at (1,4):
The equation of a tangent to a circle at a given point (x1, y1) is given by (x - x1)(x1 - h) + (y - y1)(y1 - k) = 0, where (h,k) is the center of the circle.
Substituting the values we found, the equation of the tangent at (1,4) would be:
(x - 1)(1 - 2) + (y - 4)(4 - 1) = 0
-x + y - 2 = 0
(iv) The equation of the circle concentric with the above circle and passing through the point (7,8):
To find the equation of the circle concentric with the above circle, we need to determine its radius, which can be calculated using the distance formula between the center (2,1) and the point (7,8):
r = √((x2 - x1)^2 + (y2 - y1)^2)
r = √((7 - 2)^2 + (8 - 1)^2)
r = √(5^2 + 7^2)
r = √74
Therefore, the equation of the circle concentric with the above circle and passing through the point (7,8) would be:
(x - 2)^2 + (y - 1)^2 = (√74)^2
(x - 2)^2 + (y - 1)^2 = 74
I hope this explanation helps! Please let me know if you have any further questions.