Answer:

[tex]\boxed{\bf\:Length\:of\:chord\:intercepted \: is\: 2 \sqrt{7} \: units \: } \\ [/tex]

Step-by-step explanation:

Given equation of circle is x² + y² - 8x - 2y - 8 = 0 and the equation of line is x + y + 1 = 0.

Let assume that AB be the chord intercepted by the circle x² + y² - 8x - 2y - 8 = 0 on the line is x + y + 1 = 0.

So, let's first find the point of intersection of a line with a circle.

As it is given that equation of line is

[tex]\sf \: x + y + 1 = 0 \: \implies \sf \:y = - (x + 1) \\ [/tex]

On substituting this value of y in equation of circle, we get

[tex]\sf \: {x}^{2} + {[ - (x + 1)]}^{2} - 8x + 2(x + 1) - 8 = 0 \\ [/tex]

[tex]\sf \: {x}^{2} + {x}^{2} + 2x + 1 - 8x + 2x + 2 - 8 = 0 \\ [/tex]

[tex]\sf \: 2{x}^{2} - 4x - 5 = 0 \\ [/tex]

can be rewritten as

[tex]\sf \: 2{x}^{2} - 4x + 2 - 7 = 0 \\ [/tex]

[tex]\sf \: 2({x}^{2} - 2x + 1) - 7 = 0 \\ [/tex]

[tex]\sf \: 2 {(x - 1)}^{2} = 7 \\ [/tex]

[tex]\sf \: {(x - 1)}^{2} = \dfrac{ 7 }{2} \\ [/tex]

[tex]\sf \: {(x - 1)}^{2} = \dfrac{ 14 }{4} \\ [/tex]

[tex]\sf \: x - 1 = \pm \: \dfrac{ \sqrt{14} }{2} \\ [/tex]

[tex]\sf \: x = 1 \: \pm \: \dfrac{ \sqrt{14} }{2} \\ [/tex]

[tex]\sf \: x = \dfrac{ 2 \pm\sqrt{14} }{2} \\ [/tex]

[tex]\implies \sf \: x = \dfrac{ 2 + \sqrt{14} }{2} \: \: or \: \: x = \dfrac{ 2 - \sqrt{14} }{2} \\ [/tex]

On substituting the value of x in equation (1), we get the point of intersection of a line and a circle

[tex]\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y = - (x + 1) \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf \dfrac{ 2 + \sqrt{14} }{2} & \sf \dfrac{ - (4 + \sqrt{14} )}{2} \\ \\ \sf \dfrac{ 2 - \sqrt{14} }{2} & \sf \dfrac{ - ( 4 - \sqrt{14}) }{2} \end{array}} \\ \end{gathered}[/tex]

Thus,

[tex]\sf \: Coordinates\:of\:A = \left(\dfrac{ 2 - \sqrt{14} }{2}, \: \dfrac{ - (4 - \sqrt{14}) }{2}\right) \\ [/tex]

and

[tex]\sf \: Coordinates\:of\:B = \left(\dfrac{ 2 + \sqrt{14} }{2}, \: \dfrac{ - (4 + \sqrt{14}) }{2}\right) \\ [/tex]

Now,

[tex]\sf \: Length\:of\:chord\:intercepted \\ [/tex]

[tex]\sf \: = \: AB \\ [/tex]

[tex]\sf \: = \: \sqrt{ {\left(\dfrac{ 2 + \sqrt{14} }{2} - \dfrac{ 2 - \sqrt{14} }{2}\right)}^{2} + {\left(\dfrac{ - 4 - \sqrt{14} }{2} - \dfrac{ - 4 + \sqrt{14} }{2}\right)}^{2} } \\ [/tex]

[tex]\sf \: = \: \sqrt{ {\left(\dfrac{ 2 + \sqrt{14} - 2 + \sqrt{14} }{2}\right)}^{2} + {\left(\dfrac{ - 4 - \sqrt{14} + 4 - \sqrt{14} }{2} \right)}^{2} } \\ [/tex]

[tex]\sf \: = \: \sqrt{ {\left(\dfrac{ 2\sqrt{14} }{2}\right)}^{2} + {\left(\dfrac{ - 2\sqrt{14} }{2} \right)}^{2} } \\ [/tex]

[tex]\sf \: = \: \sqrt{ {\left(\sqrt{14}\right) }^{2} + {\left( - \sqrt{14}\right) }^{2} } \\ [/tex]

[tex]\sf \: = \: \sqrt{ {\left(\sqrt{14}\right) }^{2} + {\left( \sqrt{14}\right) }^{2} } \\ [/tex]

[tex]\sf \: = \: \sqrt{2 {\left(\sqrt{14}\right) }^{2} } \\ [/tex]

[tex]\sf \: = \: \sqrt{2 \times 14 } \\ [/tex]

[tex]\sf \: = \: \sqrt{2 \times 2 \times 7 } \\ [/tex]

[tex]\sf \: = \: 2 \sqrt{7} \\ [/tex]

Hence,

[tex]\implies \sf \:\boxed{\sf\:Length\:of\:chord\:intercepted \: is\: 2 \sqrt{7} \: units \: } \\ [/tex]