Question

centre of circle q is on y axis and the circle passes through the points (0, 7) and (0,-1) .circle intersects the positive X axis at (P,0).what is the value of p ?​

Answer 1

Answer:

The center of the circle lies on y axis.  

Two other points A(0,7) and B(0,-1) also lie on y-axis.

⟹ This means that AB is the diameter of the circle.

∴Midpoint of AB = Center of the circle = (xo,yo)=(0,3)

Radius of the circle = OA = OB = 4

Equation of the circle : (x−xo)2+(y−yo)2=r2

⟹(x−0)2+(y−3)2=42⟹x2+(y−3)2=16

P = Positive X-axis intersect of the circle = (b,0)

Substituting in the equation of the circle:  

b2+(0−3)2=16⟹b2=7⟹b=±√ 7

Considering the positive value: P=(7–√,0)

OR  

So centre of the circle is ((x1+x2)/2,(y1+y2)/2 )

= ( (0+0)/2,(7–1)/2) = (0,3) and the radius of the circle is half of the distance between (0,7) and (0,-1) ie r = 8/2 = 4 units.

So equation of the circle with centre (h,k)= (0,3) and radius r= 4 is

(x-h)^2 + (y-k)^2 = r^2 => x^2 + (y-3)^2= 16

If (p,0) lies on the circle it has to satisfy the above equation.

So p^2 +(0–3)^2= 16=> p^2+9=16 => p^2=16–9

=> p^2=7 => p = + or - √7. But p is positive.

So p = √7

Step-by-step explanation:

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