Question

The length of the tangent from point (5, 1) to
the circle x2 + y2 + 6x - 4y - 3 = 0 is​

Answer 1

EXPLANATION.

The length of the tangent from the point ( 5,1)

equation of circle = x² + y² + 6x - 4y - 3 = 0

$\sf : \implies \: formula \: of \: length \: tangent \\ \\ \sf : \implies \: t \: = \sqrt{ s_{1} } \\ \\ \sf : \implies \: put \: the \: points \: in \: equation \: of \: circle$

$\sf : \implies \: t \: = \sqrt{ {5}^{2} + {1}^{2} + 6(5) - 4(1) - 3} = 0 \\ \\ \sf : \implies \: \sqrt{25 + 1 + 30 - 4 - 3} = 0 \\ \\ \sf : \implies \: \sqrt{56 - 7} = 0 \\ \\ \sf : \implies \: \sqrt{49} = 7$

$\sf : \implies \: \green{{ \underline{length \: of \: tangent \: from \: point \: (5,1) \: is \: = 7}}}$

Answer 2

✒️GIVEN:

• Length of the tangent from the point (5,1)
• The eq. of the Circle x² + y² + 6x - 4y - 3 = 0

✒️ANSWER:

7

✒️SOLUTION:

$\rm S_{1} : {x}^{2} + {y}^{2} + 6x - 4y - 3 = 0$

Center, of the above circle, $\rm S_1$, is C(-3,2) and radius

$\rm r = \sqrt{9 + 4 + 3} = \sqrt{16} = 4$

Let, given point be P(5,1)

Now,

$\rm CP = \sqrt{64 + 1} = \sqrt{65}$

Hence, length of the required tangent is,

$\rm = \sqrt{{CP}^{2} - {r}^{2} } \\ \rm \implies \sqrt{65+ 16} \\ \rm \implies \sqrt{49} = 7$

Hence, the length of the tangent = 7