Question

The equation of the circle passing through the point (1, 1) and having two diameters along the pair of lines x² − y² − 2x + 4y − 3 = 0 is

विकल्प

• (x+2)² + (y-2)² = 4

• (x-3)² + (y-1)² = 4

• (x-1)² + (y−2)² = 1 

• (x+1)² + (y +2)² = 1

Answer 1

(x-1)² + (y−2)² = 1 

Explanation:

x² − y² − 2x + 4y − 3 = 0
∴ x² − 2x + 1 − y² + 4y − 4 = 0
∴ (x − 1)² − (y − 2)² = 0
∴ (x − 1 + y − 2) (x − 1 − y + 2) = 0
∴ (x + y − 3) (x − y + 1) = 0

∴ Equations of diameters are

x + y = 3 ...(i)
x − y = −1 ...(ii)
On solving, we get
Centre = (x, y) = (1, 2)
As point (1, 1) lies on the circle,

\[\therefore\quad\mathrm{r}=\sqrt{\left(1-1\right)^{2}+\left(2-1\right)^{2}}\]

= 1
∴ The equation of a circle is
(x − 1)² + (y − 2)² = 1