The equation of the normal of the circle ${2 x}^{2} + {2 y}^{2} - 2 x - 5 y - 7 = 0$ passing through the point $(1, 1)$ is

x + 2 y - 3 = 0

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

2 x + y - 3 = 0

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

2 x + 3 y - 5 = 0

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

none of these

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

Open in App

Solution

The correct option is A $x + 2 y - 3 = 0$
Centre of the given circle is $(\frac{1}{2}, \frac{5}{4})$ .
Normal passes through centre.
Hence we need a line which is paasing through $(\frac{1}{2}, \frac{5}{4})$ and $(1, 1)$
The slope of the line is $\frac{\frac{5}{4} - 1}{\frac{1}{2} - 1} = - \frac{1}{2}$
$\Rightarrow$ Equation of normal is: $(y - 1) = - \frac{1}{2} (x - 1)$
$\Rightarrow 2 y - 2 = - x + 1$
$\Rightarrow x + 2 y - 3 = 0$

Suggest Corrections

Similar questions

Q. The equation of the circle for which

x y + 2 y + 2 x + 4 = 0

are normals and is passing through the point

(- 1, 1)

Q. The equation of line passing through the point of intersection of the lines4x-3y-1=0 and 5x-2y-3=0 and parallel to the line 2y-3x+2=0, is

Q. Find the equation of the plane passing through the point

(1, 1 - 1)

and perpendicular to the plnes

x + 2 y + 3 z - 7 = 0

and

2 x - 3 y + 4 z = 0

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. Assertion :Statement-1: Equation of a circle through the origin and belonging to the co-axial system.of which the limiting points are

(1, 1)

and

(3, 3)

2 x^{2} + 2 y^{2} - 3 x - 3 y = 0

Reason: Equation of a circle passing through the points

(1, 1)

and

(3, 3)

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

Join BYJU'S Learning Program

The equation of the normal of the circle 2x2+2y2−2x−5y−7=0 passing through the point (1,1) is

The correct option is A x+2y−3=0Centre of the given circle is (12,54). Normal passes through centre. Hence we need a line which is paasing through (12,54) and (1,1) The slope of the line is 54−112−1=−12 ⇒ Equation of normal is: (y−1)=−12(x−1) ⇒2y−2=−x+1 ⇒x+2y−3=0

The equation of the normal of the circle ${2 x}^{2} + {2 y}^{2} - 2 x - 5 y - 7 = 0$ passing through the point $(1, 1)$ is