If $a_{1}, a_{2}, a_{3}, a_{4}$ are the magnitude of the coordinates at which the circle cuts the axes as shown in the figure. Then what the sum of x and y intercepts.

(a₁ - a₃) + (a₂ - a₄)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(a₂ + a₄) + (a₁ + a₃)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

(a₂ - a₁) + (a₃ - a₄)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

(a₂ - a₄) (a₁ - a₃)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B
(a₂ + a₄) + (a₁ + a₃)

The length of intercepts is the total length between 2 points of intersections between circle and axis.

Here for x axis it $a_{2} + a_{4}$ . For y axis its $(a_{1} + a_{3})$ adding both we get $(a_{2} + a_{4}) + (a_{1} + a_{3})$

Suggest Corrections

Similar questions

If $a_{1}, a_{2}, a_{3}, a_{4}$ are the coefficients of any four

consecutive terms in the expansion of ( $1 + x)^{n}$ , then

$\frac{a_{1}}{a_{1} + a_{2}}$ + $\frac{a_{3}}{a_{3} + a_{4}}$ =

a_{1}, a_{2}, a_{3}, a_{4}, a_{5} > 0

\frac{a_{1}}{a_{2} + a_{3}} + \frac{a_{2}}{a_{3} + a_{4}} + \frac{a_{3}}{a_{4} + a_{5}} + \frac{a_{4}}{a_{5} + a_{1}} + \frac{a_{5}}{a_{1} + a_{2}} \geq \frac{5}{2}

a_{1}, a_{2}, a_{3}, a_{4}

1 + x)^{n}

\frac{a_{1}}{a_{1} + a_{2}} + \frac{a_{3}}{a_{3} + a_{4}} =

a_{1}, a_{2}, a_{3}, a_{4}

a_{1}, a_{2}, a_{3}, . . . a_{n}

\frac{a_{1}}{a_{2} + a_{3} + . . . a_{n}}, \frac{a_{2}}{a_{1} + a_{3} + . . . a_{n}}, \frac{a_{3}}{a_{1} + a_{2} + a_{4} + . . . + a_{n}}, \frac{a_{n}}{a_{1} + a_{2} + . . . + a_{n}}

Join BYJU'S Learning Program