The circles ${x^2} + {y^2} + 2x - 2y + 1 = 0\,\,\,\,\,\,\,{\rm{and}}\,\,\,\,\,\,{x^2} + {y^2} - 2x - 2y + 1 = 0$ touch each other

JEE MATHEMATICS MCQ: CIRCLES

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The circles

{x^2} + {y^2} + 2x - 2y + 1 = 0\,\,\,\,\,\,\,\mathrm{and}\,\,\,\,\,\,{x^2} + {y^2} - 2x - 2y + 1 = 0

touch each other

View answer and explanation

Correct option: A

externally at

(0,1)

Option A (externally at (0,1)) is correct because the distance between the centers equals the sum of the radii, and the common tangent equation gives the point of contact.

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