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If the length of the tangent from (2, 5) to $x^2 + y^2 - 5x + 4y + k = 0$ is $\sqrt{37}$ units, then the value of $k$ is:

This question was previously asked in
RRB NTPC 2019 CBT 1 Question Paper (8-Mar-2021) (Shift 2)
The correct answer is
-2

Finding the Value of k for a Circle Tangent Length

The length of the tangent from an external point $(x_1, y_1)$ to a circle defined by the equation $x^2 + y^2 + 2gx + 2fy + c = 0$ is given by the formula:

Length = $\sqrt{x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c}$

Alternatively, if the circle equation is given in the form $x^2 + y^2 + Dx + Ey + F = 0$, the length is $\sqrt{x_1^2 + y_1^2 + Dx_1 + Ey_1 + F}$.

Applying the Tangent Length Formula

We are given:

  • Point $(x_1, y_1) = (2, 5)$
  • Circle equation: $x^2 + y^2 - 5x + 4y + k = 0$. Here, $D = -5$, $E = 4$, and $F = k$.
  • Length of the tangent = $\sqrt{37}$

Substitute the values into the formula:

$\sqrt{37} = \sqrt{(2)^2 + (5)^2 - 5(2) + 4(5) + k}$

Calculating the Value of k

To find $k$, we first square both sides of the equation:

$37 = (2)^2 + (5)^2 - 5(2) + 4(5) + k$

Evaluate the terms:

$37 = 4 + 25 - 10 + 20 + k$

Combine the constant terms on the right side:

$37 = (4 + 25 + 20) - 10 + k$

$37 = 49 - 10 + k$

$37 = 39 + k$

Isolate $k$:

$k = 37 - 39$

$k = -2$

Therefore, the value of $k$ is -2.

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