Question

If the straight line \[ x+by+1=0 \] is a polar with respect to the circle \[ x^2+y^2-8x+10y-8=0, \] but not a tangent and not a chord, then all the values of \(b\) lie in the interval

• A. \((-\infty,-1)\)
• B. \((-\infty,-\frac43)\cup(-\frac34,\infty)\)
• C. \((-1,7)\)
• D. \(\left(-\frac43,-\frac34\right)\)

Hint:

For a line \[ ax+by+c=0, \] compare its perpendicular distance from the centre with the radius: \[ \boxed{ \begin{aligned} d d=r &\Rightarrow \text{Tangent} d>r &\Rightarrow \text{Outside the circle} \end{aligned} } \]

Answer 1

The Correct Option is D

Step 1: Find the centre and radius of the circle. The given circle is \[ x^2+y^2-8x+10y-8=0. \] Hence, \[ \boxed{\text{Centre }(4,-5)} \] and \[ r=\sqrt{16+25+8}=7. \]

Step 2: Use the condition for a line to be neither a tangent nor a chord. The line is \[ x+by+1=0. \] Its perpendicular distance from the centre is \[ d = \frac{|4-5b+1|}{\sqrt{1+b^2}} = \frac{|5-5b|}{\sqrt{1+b^2}} = \frac{5|1-b|}{\sqrt{1+b^2}}. \] For the line to be neither a tangent nor a chord, it must lie completely outside the circle. Hence, \[ d>7. \] Therefore, \[ \frac{25(1-b)^2}{1+b^2}>49. \] Expanding, \[ 25(1-2b+b^2)>49+49b^2, \] \[ 24b^2+25b+12<0. \] Factoring, \[ (4b+3)(6b+8)<0. \] Thus, \[ -\frac43<b<-\frac34. \] Hence, \[ \boxed{\left(-\frac43,-\frac34\right)}. \] Therefore, the correct option is \(\boxed{(D)}\).