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.)