Question

Points P(4, 1), Q(6, 5) and R(2, 7) lie on a circle. What are the perpendicular distances of the chords PQ and QR from the centre, respectively?

• A. $\sqrt{5}units,\sqrt{5}units$
• B. $\sqrt{5}units,\sqrt{10}units$
• C. $\sqrt{10}units,\sqrt{5}units$
• D. $\sqrt{10}units,\sqrt{10}units$

Answer 1

The correct option is A $\sqrt{5}units,\sqrt{5}units$

Let C(x, y) be the centre of the circle.

Now, points P(4, 1), Q(6, 5) and R(2, 7) lie on the circle.

∴ CP = CQ = CR

Now, CP = CQ

$\Rightarrow C{P}^2=C{Q}^2$

$\Rightarrow (4-x{)}^2+(1-y{)}^2=(6-x{)}^2+(5-y{)}^2$

$\Rightarrow 16+x^2-8x+1+y^2-2y=36+x^2-12x+25+y^2-10y$

$\Rightarrow 17-8x-2y=61-12x-10y$

$\Rightarrow 4x+8y=44$

$\Rightarrow x+2y=11$ ...(i)

Also, CP = CR

$\Rightarrow C{P}^2=C{R}^2$

$\Rightarrow (4-x{)}^2+(1-y{)}^2=(2-x{)}^2+(7-y{)}^2$

$\Rightarrow 16+x^2-8x+1+y^2-2y=4+x^2-4x+49+y^2-14y$

$\Rightarrow 17-8x-2y=53-4x-14y$

$\Rightarrow 4x-12y=-36$

$\Rightarrow x-3y=-9$ ...(ii)

Solving (i) and (ii), we get x = 3, y = 4.

Therefore, the centre of the circle is C(3, 4).

It is known that the perpendicular drawn from the centre to the chord bisects the chord.

$\therefore \text{Mid-point of PQ}=(\frac{4+6}{2},\frac{1+5}{2})=(5,3)$

Mid-point of QR $=(\frac{6+2}{2},\frac{5+7}{2})=(4,6)$

∴ Perpendicular distance from centre to chord PQ $=\sqrt{(5-3{)}^2+(3-4{)}^2}=\sqrt{4+1}=\sqrt{5}units$

Perpendicular distance from centre to chord QR $=\sqrt{(4-3{)}^2+(6-4{)}^2}=\sqrt{1+4}=\sqrt{5}units$

Hence, the correct answer is option (a).