Question

show that the tangent at (3,-2) of the circle x2+y2=13touches the circle x2+y2+2x​

Answer 1

Step-by-step explanation:

The circle x² + y² = 13 has its centre at (0,0) and radius sqrt(13). The slope of the line through the centre (0,0) and the tangent point (3,-2) = (-2 - 0)/(3 - 0) = -2/3 Since the tangent line to the circle and the line passing through the centre are perpendicular to each other, then the slope of the tangent line = 3/2. So the equation of the tangent line is:

(y + 2)/(x - 2) = 3/2 Simplify, you get: y = (3x - 13)/2 … (1) Now, to prove that this line is tangent to the circle of the equation x2 + y2 + 2x - 10y - 26 = 0 … (2) there must be one root of the quadratic equation obtained from solving (1) and (2) simultaneously, let us solve them by substituting (1) in (2):

x^2 + (9x^2 - 78x + 169)/4 + 2x - 10[(3x - 13)/2] - 26 = 0 Simplify, you get:

x^2 - 10x + 25 = 0 or (x - 5)^2 = 0 as such, x = 5 Substitute in (1) to get y:

y = (15 - 13)/2 = 2 Therefore, the tangent point between the line y = (3x - 13)/2 and the circle x2 + y2 + 2x - 10y - 26 = 0 is (5,2

Answer 2

Answer:

Tangent at (3,-2) to

${x}^{2} + {y}^{2} = 13 \: is \: 3x - 2y = 13$

now 2nd circle is

${x}^{2} + {y}^{2} + 2x = 0 \: i.e \: \: (x + 1) {}^{2} + {y}^{2} = 1$

it's center is ( -1 , 0) & Radius= 1

now, distance of (-1,0) from tangent 3x-2y-13=0 is,

$\frac{ | - 3 -0 - 13| }{ \sqrt{ {3}^{2} + {2}^{2} } } = \frac{16}{ \sqrt{13} }$

which is not equal to Radius so can't touch.

I think 2nd circle eq not written completely.