JEE MATHEMATICS MCQ: CIRCLES

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Application +4 | -1

(\alpha, \beta)

are the roots of the equation

15 x^2-22 x

+8=0

and

\left(\alpha^{\prime}, \beta^{\prime}\right)

are the roots of the equation

8 x^2-22 x+15=0

, then the equation of the circle on

A\left(\alpha, \alpha^{\prime}\right)

and

B\left(\beta, \beta^{\prime}\right)

as diameter is

View answer and explanation

Correct option: A

120\left(x^2+y^2\right)-22(8 x+15 y)+289=0

Option B (

120(x^2+y^2)-22(15 y+8 x)+289=0

) is correct because the equation of the circle with the given diameter is found by using the roots of the two quadratic equations.

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