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

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)² + (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

x² + y² – ax – by = 0

The equation of the required circle is x² + y² – ax – by = 0