Two circle of equal radius r cut orthogonally. If their centres are (2,3) and (5,6) then find value of r.

Question

Two circle of equal radius r cut orthogonally. If their centres are (2,3) and (5,6) then find value of r. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/8244f62b241d370b54cbfb6bed0beaed-.png" /> orthogonal intersection r 12+ r 22= d 2 r 2+ r 2=(5-2)2+(6-3)2 2 r 2=9+9 r 2=(18/2) ⇒ r 2=9 ⇒ r =3

Q. Two circle of equal radius $r$ cut orthogonally. If their centres are $(2, 3)$ and $(5, 6)$ then find value of $r$ .

orthogonal intersection
$r_{1}^{2} + r_{2}^{2} = d^{2}$
$r^{2} + r^{2} = (5 - 2)^{2} + (6 - 3)^{2}$
$2 r^{2} = 9 + 9$
$r^{2} = \frac{18}{2}$
$\Rightarrow r^{2} = 9$
$\Rightarrow r = 3$

Q. Two circle of equal radius r cut orthogonally. If their centres are (2,3) and (5,6) then find value of r.

orthogonal intersection r12​+r22​=d2 r2+r2=(5−2)2+(6−3)2 2r2=9+9 r2=218​ ⇒r2=9 ⇒r=3

Q. Two circle of equal radius $r$ cut orthogonally. If their centres are $(2, 3)$ and $(5, 6)$ then find value of $r$ .

orthogonal intersection
$r_{1}^{2} + r_{2}^{2} = d^{2}$
$r^{2} + r^{2} = (5 - 2)^{2} + (6 - 3)^{2}$
$2 r^{2} = 9 + 9$
$r^{2} = \frac{18}{2}$
$\Rightarrow r^{2} = 9$
$\Rightarrow r = 3$