Question

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

Answer 1

Step-by-step explanation:

Equation of a circle is

x^2+y^2=r^2

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Answer 2

$\huge{\underline{\rm{\green{\bf{Solution:-}}}}}$

Let us consider the equation of the required circle be $\sf (x - h)^{2}+ (y - k)^{2} =r^{2}$

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

According to the question,

$\sf (0 - h)^{2}+ (0 - k)^{2} = r^{2}$

$\sf h^{2} + k^{2} = r^{2}$

Now, The equation of the circle is $\sf (x - h)^{2} + (y - k)^{2} = h^{2} + k^{2}$

Given that, the circle intercepts a and b on the coordinate axes.

The circle passes through points (a, 0) and (0, b).

Hence,

$\sf (a - h)^{2}+ (0 - k)^{2} =h^{2} +k^{2} \qquad ...(1)$

$\sf (0 - h)^{2}+ (b- k)^{2} =h^{2} +k^{2} \qquad ...(2)$

From equation (1),

$\implies \sf a^{2} - 2ah + h^{2} +k^{2} = h^{2} +k^{2}$

$\implies \sf a^{2} - 2ah = 0$

$\implies \sf a(a - 2h) =0$

$\implies \sf a = 0 \: or\: (a -2h) = 0$

However, a ≠ 0; hence, (a – 2h) = 0

$\implies \sf h=\dfrac{a}{2}$

From equation (2),

$\implies \sf h^{2} - 2bk + k^{2} + b^{2}= h^{2} +k^{2}$

$\implies \sf b^{2} - 2bk = 0$

$\implies \sf b(b- 2k) = 0$

$\implies \sf b= 0 \: or \: (b-2k) =0$

However, a ≠ 0; hence, (b – 2k) = 0

$\implies \sf k=\dfrac{b}{2}$

Then, the equation is,

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

$\implies \sf \bigg[ \dfrac{(2x-a)}{2} \bigg]^{2}+ \bigg[ \dfrac{(2y+b)}{2} \bigg]^{2}=\dfrac{(a^{2}+b^{2})}{4}$

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

$\implies \sf 4x^{2} + 4y^{2} - 4ax - 4by = 0$

$\implies \sf 4(x^{2} +y^{2} -7x + 5y - 14) = 0$

$\implies \sf x^{2} + y^{2} - ax - by = 0$

∴ The equation of the required circle is $\sf \bf{x^{2} + y^{2} - ax - by = 0}$