Find the equation of the circle passing through (0,0) and making intercepts a and b on the coordinate axes.

Asked by Abhisek | 1 year ago |  69

##### Solution :-

Let us consider the equation of the required circle be (x – h)2+ (y – k)2 =r2

We know that the circle passes through (0, 0),

So, (0 – h)2+ (0 – k)2 = r2

h2 + k2 = r2

Now, The equation of the circle is (x – h)+ (y – k)2 = h2 + k2.

It is given that the circle intercepts a and b on the coordinate axes.

i.e., the circle passes through points (a, 0) and (0, b).

So, (a – h)2+ (0 – k)2 =h2 +k2……………..(1)

(0 – h)2+ (b– k)2 =h2 +k2………………(2)

From equation (1), we obtain

a2 – 2ah + h2 +k2 = h2 +k2

a2 – 2ah = 0

a(a – 2h) =0

a = 0 or (a -2h) = 0

However, a ≠ 0; hence, (a -2h) = 0

h = $$\dfrac{a}{2}$$

From equation (2), we obtain

h2 – 2bk + k2 + b2= h2 +k2

b2 – 2bk = 0

b(b– 2k) = 0

b= 0 or (b-2k) =0

However, a ≠ 0; hence, (b -2k) = 0

k =$$\dfrac{b}{2}$$

So, the equation is

(x – $$\dfrac{a}{2}$$)2 + (y – $$\dfrac{b}{2}$$)2 = ($$\dfrac{a}{2}$$)2 + ($$\dfrac{b}{2}$$)2

$$(\dfrac{2x-a}{2})^2+(\dfrac{2y-b}{2})^2=\dfrac{a^2+b^2}{4}$$

4x2 – 4ax + a2 +4y2 – 4by + b2 = a2 + b2

4x2 + 4y2 -4ax – 4by = 0

4(x2 +y2 -7x + 5y – 14) = 0

x+ y– ax – by = 0

The equation of the required circle is x+ y– ax – by = 0

Answered by Pragya Singh | 1 year ago

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