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Question 24 of 47

Q.Find the length of the tangent from the point (2,3)(2, 3) to the circle x2+y2+8x+4y+8=0x^2 + y^2 + 8x + 4y + 8 = 0.

Puducherry TnboardTamil Nadu HSC First Year (DGE) Commerce Board 2020Subjective· 2mImportance★★★★★
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Substituting (2,3)(2,3) into the circle's expression gives S1=49S_1=49, so the tangent length is 49=7\sqrt{49}=7 units.

For a circle S≡x2+y2+2gx+2fy+c=0S \equiv x^2+y^2+2gx+2fy+c=0, the length of the tangent from an external point (x1,y1)(x_1,y_1) is:

Length=S1,S1=x12+y12+2gx1+2fy1+c.\text{Length} = \sqrt{S_1},\qquad S_1 = x_1^2+y_1^2+2gx_1+2fy_1+c.

Here the circle is x2+y2+8x+4y+8=0x^2+y^2+8x+4y+8=0 and the point is (x1,y1)=(2,3)(x_1,y_1)=(2,3). Substitute: …

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