Find the equation of the circle passing through (0, 0) and making | Filo
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Class 11

Math

3D Geometry

Conic Sections

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Find the equation of the circle passing through and making intercepts and on the coordinate axes.

Solution: Let the equation of the required circle be 
$$(x - h) ^{2} + (y - k)^{2} = r ^{2}$$
Since the circle passes through $$(0, 0)$$
$$(0 - h) ^{2} + ( 0 - k) ^{2} = r \displaystyle ^{2}$$
$$\displaystyle \Rightarrow h^{2}+k^{2}= r^{2}$$
So, the equation of the circle becomes 
$$(x - h) ^{2} + (y - k)\ ^{2} = h^{2} + k ^{2}$$

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
$$\displaystyle a^{2} - 2ah + h ^{2} + k^{2} = h \displaystyle ^{2} + k^{2}$$
$$\displaystyle \Rightarrow a^{2}-2ah = 0$$
$$\displaystyle \Rightarrow  a(a - 2h) = 0$$
$$\displaystyle \Rightarrow a = 0 or (a - 2h) = 0$$
But since $$\displaystyle a\neq  0$$
Hence $$(a - 2h) = 0 \Rightarrow a=2h$$
$$\displaystyle \Rightarrow  h =  \frac{a}{2}$$

From equation (2) we obtain,
$$\displaystyle h^{2} + b^{2}  - 2bk +k ^{2}  = h^{2} + k^{2}$$ 
$$\displaystyle \Rightarrow b ^{2} - 2bk = 0$$
$$\displaystyle \Rightarrow b (b - 2k) = 0$$
$$\displaystyle \Rightarrow  b = 0 or (b - 2k) = 0$$
But since $$\displaystyle b\neq  0$$
Hence $$(b - 2k) = 0 \Rightarrow b=2k$$
$$\displaystyle \Rightarrow  k = \cfrac{b}{2}$$
Thus the equation of the required circle is
$$\displaystyle \left ( x-\frac{a}{2} \right )^{2}+\left ( y-\frac{b}{2} \right )^{2}=\left ( \frac{a}{2} \right )^{2}+\left ( \frac{b}{2} \right )^{2}$$

$$\displaystyle \Rightarrow \left ( \frac{2x - a}{2} \right )^{2}+\left ( \frac{2y - b}{2} \right )^{2}= \frac{a^{2}+b^{2}}{4}$$

$$\displaystyle \Rightarrow 4x^{2}-4ax+ a^{2}+4y^{2}-4by + b^{2}= a^{2}+b^{2}$$
$$\displaystyle \Rightarrow 4x^{2}+4y^{2}- 4ax-4by = 0$$
$$\displaystyle \Rightarrow x^{2}+y^{2}-ax - by = 0$$
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