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प्रश्न
Find the centre of the circle passing through (6, -6), (3, -7) and (3, 3)
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उत्तर
The distance d between two points `(x_1, y_1)` and `(x_2, y_2)` is given by the formula
`d = sqrt((x_1 - x_2)^2 + (y_1 - y_2)^2)`
The centre of a circle is at equal distance from all the points on its circumference.
Here it is given that the circle passes through the points A(6,−6), B(3,−7) and C(3,3).
Let the centre of the circle be represented by the point O(x, y).
So we have AO = BO = CO
`AO = sqrt((6 - x)^2 + (-6-y)^2)`
`BO = sqrt((3 - x)^2 + (-7 - y)^2)`
`CO = sqrt((3 - x)^2 + (3 - y)^2)`
Equating the first pair of these equations we have,
AO = BO
`sqrt((6 - x)^2 + (-6-y)^2) = sqrt((3 -x)^2 + (3 -y)^2)`
Squaring on both sides of the equation we have,
`(6 - x)^2 + (-6-y)^2 = (3 - x)^2 + (3 - y)^2`
`36 + x^2 - 12x + y^2 + 12y = 9 + x^2 - 6x + y^2 - 6y`
6x - 18y = 54
`x - 3y= 9`
Now we have two equations for ‘x’ and ‘y’, which are
3x + y = 7
x - 3y = 9
From the second equation we have y = 3x + 7. Substituting this value of ‘y’ in the first quation we have,
`x - 3(-3x + 7) = 9`
x + 9x - 21 = 9
10x = 30
x = 3
Therefore the value of ‘y’ is,
y = 3x + 7
= -3(3) + 7
y = -2
Hence the co-ordinates of the centre of the circle are (3, -2).
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संबंधित प्रश्न
If A(5, 2), B(2, −2) and C(−2, t) are the vertices of a right angled triangle with ∠B = 90°, then find the value of t.
Prove that the points (–3, 0), (1, –3) and (4, 1) are the vertices of an isosceles right angled triangle. Find the area of this triangle
If P (2, – 1), Q(3, 4), R(–2, 3) and S(–3, –2) be four points in a plane, show that PQRS is a rhombus but not a square. Find the area of the rhombus
Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(- 1, - 2), (1, 0), (- 1, 2), (- 3, 0)
Using the distance formula, show that the given points are collinear:
(-1, -1), (2, 3) and (8, 11)
A(-2, -3), B(-1, 0) and C(7, -6) are the vertices of a triangle. Find the circumcentre and the circumradius of the triangle.
Show that the points (2, 0), (–2, 0), and (0, 2) are the vertices of a triangle. Also, a state with the reason for the type of triangle.
Find the distance between the points (a, b) and (−a, −b).
In what ratio does the point P(−4, y) divides the line segment joining the points A(−6, 10) and B(3, −8)? Hence find the value of y.
The distance between the points (3, 1) and (0, x) is 5. Find x.
A point A is at a distance of `sqrt(10)` unit from the point (4, 3). Find the co-ordinates of point A, if its ordinate is twice its abscissa.
Show that (-3, 2), (-5, -5), (2, -3) and (4, 4) are the vertices of a rhombus.
Give the relation that must exist between x and y so that (x, y) is equidistant from (6, -1) and (2, 3).
Show that the points (a, a), (-a, -a) and `(-asqrt(3), asqrt(3))` are the vertices of an equilateral triangle.
Show that the points (2, 0), (– 2, 0) and (0, 2) are vertices of a triangle. State the type of triangle with reason
Case Study -2
A hockey field is the playing surface for the game of hockey. Historically, the game was played on natural turf (grass) but nowadays it is predominantly played on an artificial turf.
It is rectangular in shape - 100 yards by 60 yards. Goals consist of two upright posts placed equidistant from the centre of the backline, joined at the top by a horizontal crossbar. The inner edges of the posts must be 3.66 metres (4 yards) apart, and the lower edge of the crossbar must be 2.14 metres (7 feet) above the ground.
Each team plays with 11 players on the field during the game including the goalie. Positions you might play include -
- Forward: As shown by players A, B, C and D.
- Midfielders: As shown by players E, F and G.
- Fullbacks: As shown by players H, I and J.
- Goalie: As shown by player K.
Using the picture of a hockey field below, answer the questions that follow:
The point on x axis equidistant from I and E is ______.
∆ABC with vertices A(–2, 0), B(2, 0) and C(0, 2) is similar to ∆DEF with vertices D(–4, 0), E(4, 0) and F(0, 4).
If (a, b) is the mid-point of the line segment joining the points A(10, –6) and B(k, 4) and a – 2b = 18, find the value of k and the distance AB.
