JEE MATHEMATICS MCQ: CIRCLES

1 Medium

1 of 1

Application +4 | -1

The radical centre of three circles, described on the three sides

3 x-2 y+10=0, x-y+5=0

and

2 x+3 y-3=0

of a triangle as diameter, is

View answer and explanation

Correct option: B

\left(-\frac{24}{13}, \frac{29}{13}\right)

Option B (

\left(-\frac{{24}}{{13}}, \frac{{29}}{{13}}\right)

) is correct because the radical center of the three circles is the orthocenter of the triangle, which is the intersection of two altitudes.

More Questions from CIRCLES

Practice All Chapter MCQs →

{\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