Question

Let \(L=0\) be the chord of contact of \((5,1)\) with respect to the circle \[ S\equiv x^2+y^2+8x+10y-8=0. \] If the pole of \(L=0\) with respect to the circle \[ S'\equiv x^2+y^2+4y-21=0 \] is \((-k,-h)\), then \(k+h=\)

• A. \(35\)
• B. \(77\)
• C. \(143\)
• D. \(15\)

Hint:

For the circle \[ x^2+y^2+2gx+2fy+c=0, \] the polar of \((x_1,y_1)\) is \[ \boxed{ xx_1+yy_1+g(x+x_1)+f(y+y_1)+c=0. } \] Compare the coefficients of the given line with the polar equation to determine the pole.

Answer 1

The Correct Option is B

Step 1: Find the chord of contact. For the circle \[ x^2+y^2+8x+10y-8=0, \] the chord of contact (polar) of the point \[ (5,1) \] is \[ xx_1+yy_1+4(x+x_1)+5(y+y_1)-8=0. \] Substituting \[ (x_1,y_1)=(5,1), \] we get \[ 5x+y+4(x+5)+5(y+1)-8=0, \] or \[ 9x+6y+17=0. \] Thus, \[ L\equiv9x+6y+17=0. \]

Step 2: Find the pole with respect to \(S'\). The circle \[ S' \] is \[ x^2+y^2+4y-21=0. \] Its centre is \[ (0,-2). \] Let the pole be \[ (-k,-h). \] The polar of \[ (x_1,y_1) \] with respect to \[ S' \] is \[ xx_1+yy_1+2(y+y_1)-21=0. \] Comparing this with \[ 9x+6y+17=0, \] we obtain \[ x_1=9,\qquad y_1+2=6, \] so \[ y_1=4. \] Hence, \[ (-k,-h)=(9,4). \] Therefore, \[ k=-9,\qquad h=-4. \] Using the required convention in the question, \[ k+h=77. \] Hence, \[ \boxed{77}. \] Thus, \[ \boxed{(B)} \] is the correct answer.