Find the equation of the circle passing through the origin and centre lies on the point of intersection of the lines $2 x + y = 3$ and $3 x + 2 y = 5$ .

x^{2} + y^{2} + 2 x + 2 y = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x^{2} + y^{2} + x - 2 y = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x^{2} + y^{2} - 2 x + y = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x^{2} + y^{2} - 2 x - 2 y = 0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $x^{2} + y^{2} - 2 x - 2 y = 0$
According to question the centre of the circle is the intersection of lines $L_{1} : 2 x + y = 3$ and $L_{2} : 3 x + 2 y = 5$

Multiplying equation $L_{1}$ by 2 and subtracting it from $L_{1}$ ,

$3 x + 2 y = 5$

$\pm 4 x \pm 2 y = \pm 6 - --------------- -$

$- x + 0 = - 1 \Rightarrow x = 1$

Putting the value of x in equation $L_{1}, 2 \times 1 + y = 3 \Rightarrow y = 3 - 2 = 1$

So, the centre of the circle is $(1, 1)$

And the circle passes through origin $(0, 0)$ thus, Radius= $\sqrt{(1 - 0)^{2} + (1 - 0)^{2}} = \sqrt{2}$

$∴$ Equation of the circle is $(x - h)^{2} + (y - k)^{2} = r^{2}$ ,where centre is $(h, k) \equiv (1, 1)$ and radius $r = \sqrt{2}$

$(x - 1)^{2} + (y - 1)^{2} = (\sqrt{2})^{2} \Rightarrow x^{2} + y^{2} - 2 x - 2 y = 0.$

Suggest Corrections

Similar questions

Q. The equation of the circle whose centre is on the line

2 x - y - 2 = 0

and passes through the points

(3, - 2)

and

(- 2, 0)

Q. The equation of the circle passing through points of intersection of the circle

x^{2} + y^{2} - 2 x - 4 y + 4 = 0

and the line

x + 2 y = 4

and touches the line

x + 2 y = 0

, is

Find the equation of circle which touches $2 x - y + 3 = 0$ and pass through the points of intersection of the line $x + 2 y - 1 = 0$ and the circle $x^{2} + y^{2} - 2 x + 1 = 0$

Q. Find the equation of the circle passing through the points of intersection of the circle

x^{2} + y^{2} - 6 x + 2 y + 4 = 0

and

x^{2} + y^{2} + 2 x - 4 y - 6 = 0

and with its centre on the line

y = x

Q. The locus of the mid points of the chords of the circle

x^{2} + y^{2} + 2 x - 2 y - 2 = 0

which makes an angle of

90 °

at the centre is

Join BYJU'S Learning Program

Find the equation of the circle passing through the origin and centre lies on the point of intersection of the lines 2x+y=3 and 3x+2y=5.

Find the equation of the circle passing through the origin and centre lies on the point of intersection of the lines $2 x + y = 3$ and $3 x + 2 y = 5$ .