Find the length of the tangent from the point (2, – Class 11 Business Mathematics Statistics | ConceptFirst

Question 24 of 47

(2, 3)

Tamil Nadu DgeTamil Nadu HSC First Year (DGE) Commerce Board 2020Subjective· 2mImportance★★★★★

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Substituting $(2,3)$ into the circle's expression gives $S_1=49$ , so the tangent length is $\sqrt{49}=7$ units.

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

$\text{Length} = \sqrt{S_1},\qquad S_1 = x_1^2+y_1^2+2gx_1+2fy_1+c.$

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

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2022Set MARCH1 markMCQ
$(1,\ -2)$
(a) $4$
(b) $3$
(c) $1$
(d) $2$
›Reveal solutionSolution

Match the centre $\left(-\tfrac a2,-\tfrac b2\right)=(1,-2)$ to get $a=-2,\ b=4$ , then radius $=\sqrt{1+4+4}=3$ .

Compare with the general circle $x^2+y^2+2gx+2fy+c=0$ whose centre is $(-g,-f)$ and radius $\sqrt{g^2+f^2-c}$ . Here $2g=a$ and $2f=b$ , so the centre is $\left(-\tfrac a2,-\tfrac b2\right)$ .

Given centre $(1,-2)$ :

…

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CBSE 2018Set ANNUAL1 markMCQ
$Q. If the circle has both x and y axes as tangents and has radius 1 unit then the equation of the circle is:$
(a) $(x+1)^2 + (y+1)^2 = 1$
(b) $x^2 + (y-1)^2 = 1$
(c) $(x-1)^2 + y^2 = 1$
(d) $x^2 + y^2 = 1$
›Reveal solutionSolution
Tangency to the x-axis and y-axis both require the center's distance to each axis to equal the radius (1); only the circle centered at $(-1,-1)$ satisfies both.

A circle with center $(h,k)$ and radius $r$ is tangent to the x-axis when $|k|=r$ , and tangent to the y-axis when $|h|=r$ . Here $r=1$ , so we need $|h|=1$ and $|k|=1$ , i.e. center at $(\pm1,\pm1)$ .

$(x+1)^2+(y+1)^2=1$ : center $(-1,-1)$ , distances to both axes $=1=r$ . Tangent to both. Valid.

$x^2+(y-1)^2=1$ : center $(0,1)$ , distance to y-axis $=0\ne1$ — the circle crosses the y-axis, not tangent. Invalid. …
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