JEE MATHEMATICS MCQ: CIRCLES

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Application +4 | -1

Find the equation of the normal to the circle

x^2+y^2-5 x+2 y

-48=0

at the point

(5,6)

View answer and explanation

Correct option: A

14 x-5 y-40=0

Option A (

14 x-5 y-40=0

) is correct because it represents the equation of the normal to the circle at the given point (5,6), which is perpendicular to the tangent at that point.

More Questions from CIRCLES

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The sum of the coefficients in the expansion of

{\left( {{\alpha ^2}{x^2} - 2\alpha \mathrm{ }x + 1} \right)^{51}},

as a polynomial in

x

, vanishes. Position of the point

\left( {\alpha ,2{\alpha ^2}} \right)

with respect to the circle

{x^2} + {y^2} = 4,

is One of the diameter of the circle

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

is The locus of the middle point of the chords of the circle

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

such that the chords pass through a given point

\left( {{x_1},{y_1}} \right)

, is

\quad \alpha, \beta

and

\gamma

are the parametric angles of three points

P, Q

, and R respectively, on the circle

x^2+y^2=1

, and

A

is the point

(-1,0)

. If the lengths of the chords

A P, A Q

and

A R

are in

G P

, then

\cos \alpha / 2, \cos \beta / 2

and

\cos \gamma / 2

are in The radius of the circle passing through the point (6,2), two of whose diameters are

x+y=6

and

x+2 y=4

is If a circle has two of its diameters along the lines

x+y=5

and

x-y=1

and has area

9 \pi

, then the equation of the circle is