Question

1.
AB is a diameter of a circle, whose centre is (3,5) and B is (6,2). find the coordinate of point A​

Answer 1

✬ B = (0 , 8) ✬

Step-by-step explanation:

Given:

• AB is diameter of circle.
• Center of AB is (3,5)
• Coordinate of B is (6,2)

To Find:

• What is the coordinate of A ?

Solution: Let the coordinate of A be (x¹ , y¹).

• A = (x¹ , y¹)

• O (centre of AB) = (3 , 5) or (a , b)

• B = (x² , y²)

Applying mid point formula

★ For x coordinate = x¹ + x²/2 ★

★ For y coordinate = y¹ + y²/2 ★

➟ a = x¹ + x²/2

➟ 3 = x¹ + 6/2

➟ 3 × 2 – 6 = x¹

➟ 0 = x¹

Now for y coordinate

➼ b = y¹ + y²/2

➼ 5 = y¹ + 2/2

➼ 5 × 2 – 2 = y¹

➼ 8 = y¹

Hence, the coordinate of point B is (0 , 8)

Answer 2

Given:-

• AB is a diameter of a circle, whose centre is (3,5) and B is (6,2).

To Find:-

• Coordinate of point A.

Theorem Used:-

• ${\boxed{\bf{\dfrac{x_1+x_2}{2},\dfrac{y_1+y_2}{2}=x,y}}}$

Solution:-

Let Point A be $\bf(x_1 , y_1)$

$\huge{\bf{\underline{For \:x_1 :}}}$

Using,

• $\sf :\implies\: x=\dfrac{x_1+x_2}{2}$

Here,

• $\bf x=3$

• $\bf x_2=6$

Putting values,

$\sf :\implies\: 3=\dfrac{x_1+6}{2}$

$\sf :\implies\: x_1+6=6$

$\sf :\implies\: x_1=6-6$

${\bf{\red{\underline{:\implies\: x_1=0}}}}$

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$\huge{\bf{\underline{For \:y_1 :}}}$

Using,

• $\sf :\implies\: y=\dfrac{y_1+y_2}{2}$

Here,

• $\bf y=5$

• $\bf y_2=2$

Putting values,

$\sf :\implies\: 5=\dfrac{y_1+2}{2}$

$\sf :\implies\: y_1+2=10$

$\sf :\implies\: y_1=10-2$

${\bf{\red{\underline{:\implies\: y_1=8}}}}$

Hence, The Coordinates of point A is (0 , 8).

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