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

If the length of the tangent from (2, 5) to the circle x²+y²-5x+4y+k=0 is √(37) then find k.

Question 2 of 9

x^2+y^2-5x+4y+k=0

Yanam BieapBIEAP Intermediate Board 2024Subjective· 2mImportance★★★★★

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The length of the tangent from $(x_1,y_1)$ to $x^2+y^2+2gx+2fy+c=0$ is $\sqrt{x_1^2+y_1^2+2gx_1+2fy_1+c}$ ; substitute and equate to $\sqrt{37}$ .

For the circle $x^2+y^2-5x+4y+k=0$ , here $2g=-5$ , $2f=4$ , $c=k$ .

Length of tangent from $(2,5)$ : …