A circle passing through the point P in the first quadrant touches the two coordinate axes at the points A and B. The point P is above the line AB. The point Q on the line segment AB is the foot of perpendicular from P on AB. If PQ is equal to 11 units,

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
A circle passing through the point $P (α, β)$ in the first quadrant touches the two coordinate axes at the points A and B. The point P is above the line AB. The point Q on the line segment AB is the foot of perpendicular from P on AB. If PQ is equal to 11 units, then the value of $α β$ is ________. [2023]

Ans.
(121)

$Let the equation of the circle be$

${(x - a)}^{2} + {(y - a)}^{2} = a^{2},$

It passes through $P (α, β)$

$∴$ ${(α - a)}^{2} + {(β - a)}^{2} = a^{2}$

$\Rightarrow α^{2} + β^{2} - 2 a α - 2 a β + a^{2} = 0$ ...(i)

Here, the equation of line AB is $x + y = a$

Let $Q (α', β')$ be the foot of the perpendicular from $P$ on AB

$∴$ $\frac{α' - α}{1} = \frac{β' - β}{1} = \frac{- (α + β - a)}{2}$

${(P Q)}^{2} = {(α' - α)}^{2} + {(β' - β)}^{2} = \frac{1}{4} {(α + β - a)}^{2} + \frac{1}{4} {(α + β - a)}^{2}$

$\Rightarrow {(11)}^{2} = \frac{1}{2} {(α + β - a)}^{2}$

$\Rightarrow 242 = α^{2} + β^{2} + a^{2} + 2 α β - 2 a α - 2 a β$

$\Rightarrow 242 = 2 α β$ [From (i)]

$\Rightarrow α β = 121$

Want Us to Email you About Special Offers & Updates?