JEE MATHEMATICS MCQ: CIRCLES

1 Hard

1 of 1

Analyzing +4 | -1

The set of values of '

c

' so that the equation

y = \left| x \right| + c

and

{x^2} + {y^2} - 8\left| x \right| - 9 = 0

have no solution, is

View answer and explanation

Correct option: D

\left( {5\sqrt 2 - 4,\;\infty } \right)

Option C is correct because eliminating the parameter 't' results in the equation of a parabola.

More Questions from CIRCLES

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

x

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

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

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

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

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

\quad \alpha, \beta

\gamma

P, Q

x^2+y^2=1

A

(-1,0)

A P, A Q

A R

G P

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

\cos \gamma / 2

x+y=6

x+2 y=4

x+y=5

x-y=1

9 \pi