Question

If the coordinates of one end of a diameter of a circle are (2, 3) and the coordinates of its centre are (-2, 5), then the coordinates of the other end of the diameter are (a)(-6, 7) (b) (6, -7) (c) (4, 2) (d) (5,3) *​

Answer 1

Answer- The above question is from the chapter 'Coordinate Geometry'.

Concept used: Distance formula-

$\sqrt{(x_{2} - x_{1})^{2} + (y_{2} -y_{1})^{2}}$

Given question: If the coordinates of one end of a diameter of a circle are (2, 3) and the coordinates of its centre are (-2, 5), then the coordinates of the other end of the diameter are

(a) (-6, 7)

(b) (6, -7)

(c) (4, 2)

(d) (5,3)

Answer: Let there be a circle with centre O and AB be its diameter.

Coordinates of O are (-2, 5).

Coordinates of A are (2, 3).

Let coordinates of B be (x,y)

We know that OA = OB (radii of circle).

By applying distance formula, we get,

$\sqrt{(-2 - 2)^{2} + (5 -3)^{2}} = \sqrt{(x +2)^{2} + (y - 5)^{2}}$

$\sqrt{(-4)^{2} + 4^{2} } = \sqrt{x^{2} + 4x + 4 + y^{2} -10y + 25$

Squaring both sides, we get,

16 + 16 = (x + 2)² + (y - 5)²

On comparison,

(x + 2)²  = 16

x + 2 = ±√16

x + 2 = ± 4

When x + 2 = + 4,

x = 4 - 2

x = 2

When x + 2 = - 4

x = - 4 - 2

x = -6

Again, on comparison,

(y - 5)² = 16

y - 5 = ±√16

y - 5 = ±4

When y - 5 = + 4

y = 4 + 5

y = 9

When y - 5 = - 4

y = - 4 + 5

y = 1

∴ Coordinates of B can be either (2 , 9) or (-6 , 1).

(None of the options above is true.)

Answer 2

Question:

If the coordinates of one end of a diameter of a circle are (2, 3) and the coordinates of its centre are (-2, 5), then the coordinates of the other end of the diameter are:

1. (-6, 7)
2. (6, -7)
3. (4, 2)
4. (5,3) 

Distance Formula:

$\red{\sqrt{{( x_{2} - x_{1}) }^{2}+{( y_{2} - y_{1}) }^{2}}}$

Let,

• Circle = O
• It's diameter = AB

1. Coordinates of O are (-2, 5)
2. Coordinates of A are (2, 3)
3. Let, coordinates of B be (x,y)

By applying the formula of distance,

$= \sqrt{{( -2 - 2) }^{2}+{( 5 - 3) }^{2}}$

$= \sqrt{{(x + 2)}^{2} + {(y - 5)}^{2}}$

$= \sqrt{{(-4)}^{2} {-4}^{2}}$

$= \sqrt{{x}^{2} + 4x + 4 + {y}^{2} - 10y + 25}$

When we squaring both sides we get,

16 + 16 = (x + 2)² + (y - 5)²

On comparison this we get,

→ (x + 2)²  = 16

→ x + 2 = ±√16

→ x + 2 = ± 4

When, x + 2 = + 4,

→ x = 4 - 2

→ x = 2

When, x + 2 = - 4

→ x = - 4 - 2

→ x = -6

Again,

On comparison this, (y - 5)² = 16

→ y - 5 = ±√16

→ y - 5 = ±4

When, y - 5 = + 4

→ y = 4 + 5

→ y = 9

When, y - 5 = - 4

→ y = - 4 + 5

→ y = 1

Hence,

The coordinates of B can be either (2 , 9) or (-6 , 1).

(None of these above option which is given in the question is true)