Question

Given point are $P=(1,-2)$, $Q=(7,6)$ is the origin. The length of the common chord of the circles with OP and OQ as diameters is

• A. $1$
• B. $2$
• C. $4$
• D. $6$

Answer 1

Answer: (D)

$6$

REF.Image.

P=(1,-2) & Q = (7,6)

distance b/w PQ = $\sqrt{(7-1{)}^2+(6+2{)}^2}=\sqrt{6^2+8^2}$

$=\sqrt{100}$

$=10$

radius of circle with centre (1,2) & point (0,0)

$(x-1{)}^2+(y+2{)}^2=r_1^2$

putting (0,0) $=1^2+2^2=r_1^2$

$1+4=r_1^2=5\Rightarrow r_1=\sqrt{5}$

lll y with centre (7, 6) of pt.(0,0)

$(x-7{)}^2+(y-6{)}^2=r_2^2$

(putting (0,0)) $7^2+(-6{)}^2=r_2^2$

$49+36=r_2^2=85\Rightarrow r_2=\sqrt{85}$

Now, the length OB i.e the common cord is given by

$=\frac{(d^2-(r_1-r_2{)}^2\left)\right((r_1+r_2{)}^2)-d^2}{d^2}$

$=\frac{\left(\right(10{)}^2)-(\sqrt{5}-\sqrt{85}{)}^2\left)\right((\sqrt{5}+\sqrt{85}{)}^2-(10{)}^2)}{(10{)}^2}$

$\Rightarrow \frac{(100-(\sqrt{5}(1-\sqrt{17}){)}^2\left)\right(\left(\sqrt{5}\right(1+\sqrt{17}\left){)}^2\right)-100}{10}$

$=\frac{(100-5(1+17-2\sqrt{17}\left)\right)\left(5\right(1+17+2\sqrt{17})-100)}{100}$

$=\frac{(10+10\sqrt{7})(10\sqrt{7}-10)}{100}.\frac{(10\sqrt{7}{)}^2-(10{)}^2}{100}$

$=\frac{100\times 7-100}{100}=\frac{600}{100}=6$

distance / length of common chord = 6