The radius of the circle whose two normals are represented by the equation x2-5 x y-5 x+25 y=0 and which touches externally the circle x2+y2-2 x+4 y-4=0 is equal to.

Question

The radius of the circle whose two normals are represented by the equation x2-5 x y-5 x+25 y=0 and which touches externally the circle x2+y2-2 x+4 y-4=0 is equal to. (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Normals are x2-5 x y-5 x+25 y=0 or (x-5)(x-5 y)=0 Hence centre of first circle is (5,1). Given circle is or (x-1)2+(y+2)2=32 Centre of this circle is (1,-2) and radius is 3 . Since circles touch externally, C1 C2=r1+r2 ⇒ 5=3+r2 ∴ r2=2

Q. The radius of the circle whose two normals are represented by the equation x2−5xy−5x+25y=0 and which touches externally the circle x2+y2−2x+4y−4=0 is equal to________.

Normals are x2−5xy−5x+25y=0 or (x−5)(x−5y)=0 Hence centre of first circle is (5,1). Given circle is or (x−1)2+(y+2)2=32 Centre of this circle is (1,−2) and radius is 3. Since circles touch externally, C1​C2​=r1​+r2​ ⇒5=3+r2​ ∴r2​=2

Q. The radius of the circle whose two normals are represented by the equation $x^{2} - 5 x y - 5 x + 25 y = 0$ and which touches externally the circle $x^{2} + y^{2} - 2 x + 4 y - 4 = 0$ is equal to________.