Question

If a circle (x+2)^2 + (y-3)^2=25 intersects, the x-axis at point A and B. Then distance of midpoint of Ab from centre of the circle will be ?​

Answer 1

Answer:

For point R,x=−

5

3

⇒y=1−

5

3m

R(−

5

3

,1−

5

3m

)

Slope of CR=

−

5

3

−3

1−

5

3m

+2

=−

m

1

⇒

−3−15

15−3m

=−

m

1

15m−3m

2

=18

m

2

−5m+6=0

m=2,3

2≤m≤4.

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Answer 2

Answer:

The distance of midpoint of AB from center of the circle is 3 units.

Step-by-step explanation:

Given equation of circle : (x + 2)² + (y - 3)² = 25

We know from the equation, the center = (-2, 3)

It is given that the circle intersects x axis at points A and B

=> The y coordinate at A and B will be 0

=> On substituting these values in the equation of circle, we get

(x + 2)² + (0 - 3)² = 25

= x² + 4 + 4x + 9 = 25

= x² + 4x - 12 = 0

= x² + 6x - 2x - 12 = 0

=> x (x + 6) - 2 (x + 6) = 0

=> (x + 6) (x - 2) = 0

=> x = -6 , x = 2

Therefore, the points of A and B are (-6, 0) and (2, 0)

The mid point of AB = $(\frac{-6+2}{2}, \frac{0+0}{0})$ = (-2, 0)

We need to find the distance of midpoint of AB from center of the circle.

Therefore, by distance formula,

The distance between (-2, 3) and (-2, 0) = $\sqrt{[-2-(-2)]^2 + (3-0)^2 }$

                                                                  = $\sqrt{[-2+2]^2 + 3^2 }$

                                                                  = $\sqrt{0^2 + 3^2 }$

                                                                  = $\sqrt{ 3^2 }$

                                                                  = 3

Therefore, the distance of midpoint of AB from center of the circle is 3 units.