Advertisements
Advertisements
प्रश्न
Find the centre of a circle passing through the points (6, − 6), (3, − 7) and (3, 3).
उत्तर
Let O (x, y) be the centre of the circle. And let the points (6, −6), (3, −7), and (3, 3) be representing the points A, B, and C on the circumference of the circle.
`:.OA = sqrt((x-6)^2+(y+6)^2)`
`OB = sqrt((x-3)^2+(y+7)^2)`
`OC = sqrt((x-3)^2+(y-3)^2)`
However OA = OB (Radii of same circle)
`=>sqrt((x-6)^2+(y+6)^2)=sqrt((x-3)^2+(y+7)^2)`
=>x2+36 - 12x + y2 + 36 + 12y = x2 + 9 -6x + y2 + 49 -14y
⇒ -6x + 2y + 14 = 0
⇒ 3x + y = 7 ....1
Similary OA = OC (Radii of same circle)
`=sqrt((x-6)^2+(y+6)^2) = sqrt((x-3)^2 + (y -3)^2)`
=x2 + 36 - 12x +y2 + 36 + 12y = x2 + 9 - 6x + y2 + 9 - 6y
⇒ -6x + 18y + 54 = 0
⇒ -3x + 9y = -27 .....(2)
On adding equation (1) and (2), we obtain
10y = −20
y = −2
From equation (1), we obtain
3x − 2 = 7
3x = 9
x = 3
Therefore, the centre of the circle is (3, −2).
APPEARS IN
संबंधित प्रश्न
If the coordinates of two points A and B are (3, 4) and (5, – 2) respectively. Find the coordniates of any point P, if PA = PB and Area of ∆PAB = 10
The area of a triangle is 5 sq units. Two of its vertices are (2, 1) and (3, –2). If the third vertex is (`7/2`, y). Find the value of y
If `a≠ b ≠ c`, prove that the points (a, a2), (b, b2), (c, c2) can never be collinear.
Prove that the points (a, b), (a1, b1) and (a −a1, b −b1) are collinear if ab1 = a1b.
In a ΔABC, AB = 15 cm, BC = 13 cm and AC = 14 cm. Find the area of ΔABC and hence its altitude on AC ?
Show that the following points are collinear:
A(5,1), B(1, -1) and C(11, 4)
Find the value of x for which the points (x, −1), (2, 1) and (4, 5) are collinear ?
A(6, 1), B(8, 2) and C(9, 4) are three vertices of a parallelogram ABCD. If E is the midpoint of DC, find the area of ∆ADE.
If `D((-1)/2, 5/2), E(7, 3)` and `F(7/2, 7/2)` are the midpoints of sides of ∆ABC, find the area of the ∆ABC.
Using determinants, find the area of ΔPQR with vertices P(3, 1), Q(9, 3) and R(5, 7). Also, find the equation of line PQ using determinants.