Find the equation of the circle which passes through the origin and cu

1.	Find the equation of the circle which passes through the origin and cuts off intercepts a and b respectively from x and y – axes.
Answer» Since the circle has intercept ‘b’ from x – axis, the circle must pass through (0, b) and (0, -b) as it already passes through the origin. Let us assume the circle passing through the points A(a,0) and B(0,b). We know that the equation of the circle with AB as diameter is given by (x – x₁) (x – x₂) + (y – y₁) (y – y₂) = 0 (x – a) (x – 0) + (y – 0) (y – b) = 0 x² + y² + ax + by = 0 or x² + y² – ax – by = 0 ∴The equation of the circle is x² + y² + ax + by = 0 or x² + y² – ax – by = 0

Find the equation of the circle which passes through the origin and cuts off intercepts a and b respectively from x and y – axes.

Answer»

Since the circle has intercept ‘b’ from x – axis, the circle must pass through (0, b) and (0, -b) as it already passes through the origin.

Let us assume the circle passing through the points A(a,0) and B(0,b).

We know that the equation of the circle with AB as diameter is given by

(x – x₁) (x – x₂) + (y – y₁) (y – y₂) = 0

(x – a) (x – 0) + (y – 0) (y – b) = 0

x² + y² + ax + by = 0 or x² + y² – ax – by = 0

∴The equation of the circle is x² + y² + ax + by = 0 or x² + y² – ax – by = 0

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Find the equation of the circle which passes through the origin and cuts off intercepts a and b respectively from x and y – axes.

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