Question

Find the equation of a circle touching the parabola y2 = 8x at (2, 4) and passes through (0,4).
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Answer 1

Answer:

(x-1)²+ (y-5)² =2

Step-by-step explanation:

Let the equation of the circle is (x-a)²+(y-b)²=r²........ (1)

Where (a,b) is the center and r is the radius of the circle.

It is clearly stated that the circle (1) passes through (2,4) and (0,4) points.

Hence, (2-a)²+(4-b)²= r²= (0-a)²+(4-b)² ....... (2)

⇒(2-a)² =a²

⇒ 4-4a=0

⇒ a= 1.

Now, given that the circle touches the parabola y²=8x ....... (3) at point (2,4).

Hence, the slope of the tangent at (2,4) to the curves (1) and (3) are the same.

Now, differentiating (1) with respect to x, we get,

2(x-a)+2(y-b)(dy/dx)=0

⇒ dy/dx=(x-a)/(b-y)

⇒[dy/dx] at (2,4)=(2-a)/(b-4) ........ (4)

Again differentiating (3) with respect to x, we get

2y(dy/dx)=8

⇒dy/dx =4/y

⇒[dy/dx] at (2,4)=4/4=1 ....... (5)

Now, equating (4) and (5), we get,

2-a=b-4

⇒a+b=6

Putting a=1, we get, b=5.

Again, from equation (2), 1²+(-1)²=r², ⇒r²=2

Hence, the equation of the circle will be  

(x-1)²+ (y-5)² =2 (Answer)