Question

$\bold\pink{Question:}$Find the equation of the circle passing through (0, 0) and making intercepts a and b on the coordinate axes.

__thank you ☺​

Answer 1

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

Since the centre of the circle passes through (0,0)

(0−h)2+(0−k)2=r2

⇒h2+k2=r2

∴ equation of the circle is

(x−h)2+(y−k)2=h2+k2

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

That is the circle passes through the points (a, 0) and (0, b)

∴ Equation of the circle is

(a−h)2+(0−k)2=h2+k2

(i.e) a2−2ah+h2+k2=h2+k2

⇒a2−2ah=0

(i.e) a(a−2h)=0

⇒a=0 or a=2h

Similarly

Equation of the circle when it passes through (0,b) is

(0−h)2+(b−k)2=h2+k2

⇒h2+b2−2bk+k2=h2+k2

(i.e) b2−2bk=0

⇒b(b−2k)=0

⇒b=0 or b=2k

But a≠0 and b≠0

∴a=2h and b=2k

or h=a2 and k=b2

is the equation of the required circle.

Answer 2

Answer:

x^2 + y^2 - ax - by = 0

Step-by-step explanation:

Let the equation of the circle be:

(x - h)² + (y - k)² = r²

Given that it passes through (0,0) and making intercepts a and b.

The centre of circle (h,k) = (a + 0/2, b + 0/2) = (a/2, b/2).

∴ Radius = √(a/2 - 0)² + (b/2 - 0)²

               = √(a² + b²)/2

Therefore, Equation of the circle is:

$\Longrightarrow(x - \frac{a}{2})^2 + (y - \frac{b}{2})^2 = (\frac{\sqrt{a^2 + b^2}}{2})^2$

$\Longrightarrow x^2 + y^2 - ax - by + \frac{a^2 + b^2}{4} = \frac{a^2 + b^2}{4}$

$\Longrightarrow \boxed{x^2+y^2-ax-by = 0}$

Hope it helps!