Question

if the length of tge tangent from (2,3) to the circle x^2+y^2+6x+2ky-6=0 is equal to 7 then k= ​

Answer 1

Step-by-step explanation:

We have,

Equation of circle:

${x}^{2} + {y}^{2} + 6x + 2ky - 6 = 0$

Differentiating both sides w.r.t x we get,

$2x + 2y \frac{dy}{dx} + 6 + 2k \frac{dy}{dx} = 0$

$\implies 2(x + y \frac{dy}{dx} + 3 + k \frac{dy}{dx} ) = 0$

$\implies(y + k) \frac{dy}{dx} + (x + 3) = 0$

$\implies \frac{dy}{dx} = - (\frac{x + 3}{y + k} )$

$\implies( \frac{dy}{dx} )_ {( \alpha \: \: and \: \: \beta )} = - \frac{ \alpha + 3 }{ \beta + k }$

so, slope of tangent is dy/dx at (α,β)

Equation of tangent

$(y - \beta ) = - \frac{ \alpha + 3}{ \beta + k } (x - \alpha )$

Since, it passes through (2,3), so we have,

$(3 - \beta ) = - \frac{ \alpha + 3}{ \beta + k} (2 - \alpha )$

$\implies(3 - \beta )( \beta + k) + ( \alpha + 3)(2 - \alpha ) = 0$

$\implies3 \beta + 3k - { \beta }^{2} - \beta k + 2 \alpha + 6 - { \alpha }^{2} - 3 \alpha = 0$

$\implies { \alpha }^{2} + \beta ^{2} + 3 \alpha - 2 \alpha + k\beta - 3 \beta - 3(k + 2) = 0$

$\implies { \alpha }^{2} + { \beta }^{2} + \alpha + (k - 3) \beta - 3(k + 2) = 0.......(i)$

Also, α & β satisfying the equation of circle

${ \alpha }^{2} + { \beta }^{2} + 6 \alpha + 2k \beta - 6 = 0.....(ii)$

Now, distance between the points (2,3) and (α,β) is 7

so,

$( \alpha - 2)^{2} + {( \beta - 3)}^{2} = 49$

$\implies { \alpha }^{2} - 4 \alpha + 4 + { \beta }^{2} - 6 \beta + 9 = 49$

$\implies { \alpha }^{2} + { \beta }^{2} - 4 \alpha - 6 \beta - 36 = 0.....(iii)$

From (i) and (ii), we get,

$5 \alpha + (k + 3) \beta + 3k = 0.....(iv)$

From (ii) and (iii), we get,

$10 \alpha + (2k + 6) \beta + 30 = 0$

$\implies5 \alpha + (k + 3) \beta + 15 = 0......(v)$

On comparing (iv) and (v), we get,

$3k = 15 \\ \implies \: k = 5$