Question

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

Answer 1

Let the equation of the required circle be \((x - h)^2 + (y - k) ^2 = r^2. \)
Since the circle passes through \((0, 0), \)
\((0 - h) ^2 + (0 - k) ^2 = r ^2 \)
\(⇒ h ^2 + k ^2 = r^2 \)
The equation of the circle now becomes \((a - h) ^2 + (0 - k)^ 2 = h ^2 + k ^2 … (1)\). 
It is given that the circle makes intercepts a and b on the coordinate axes. 
This means that the circle passes through points \( (a, 0) \)and \((0, b).\)

Therefore, \((a - h) ^2 + (0 - k)^ 2 = h ^2 + k ^2 … (1)\)
\((0 - h) ^2 + (b - k) ^2 = h ^2 + k ^2 … (2)\) 
From equation (1), we obtain
\(a ^2 - 2ah + h ^2 + k ^2 = h ^2 + k ^2 \)
\(⇒ a^2 - 2ah = 0 \)
\(⇒ a(a - 2h) = 0 \)
\(⇒ a = 0\)\(\) or \((a - 2h) = 0 \)
However,\( a ≠ 0\); hence, \((a - 2h) = 0 \)
\(⇒ h = a/2\)
From equation (2), we obtain 
\(h^2 + b^2 - 2bk + k^2 = h ^2 + k ^2 \)
\(⇒ b^2 - 2bk = 0\) 
\(⇒ b(b - 2k) = 0 \)
\(⇒ b = 0\) or \((b - 2k) = 0 \)
However, \(b ≠ 0\) ; hence, \((b - 2k) = 0 \)
\(⇒ k = b/2.\)

Thus, the equation of the required circle is

\((x – a/2)^2 + (y – b/2)^2 = (a/2)^2 + (b/2)^2\)

\([(2x-a)/2]^2 + [(2y-b)/2]^2= (a^2 + b^2)/4\)

\(4x^2 – 4ax + a^2+4y^2 – 4by + b^2 = a^2 + b^2\)

\(4x^2+ 4y^2-4ax – 4by = 0\)

\(4(x^2 +y^2-7x + 5y – 14) = 0\)

\(x^2 + y^2 – ax – by = 0\)

∴ The equation of the required circle is  \(x^2 + y^2 – ax – by = 0\)  (Ans.)