(a) Show that the circles $x^2+y^2-2x+6y+6=0$ and $x^2+y^2-5x+6y+15=0$ touch each other. **OR** (b) Evaluate: $\displaystyle\lim_{x\to 0} \dfrac{|x-1|+|x-2|-3}{2|x-1|-|x-2|}$.

(a) Show that the circles x²+y²-2x+6y+6=0 and x²+y²-5x+6y+15=0 touch each other. **OR** (b) Evaluate: limx→ 0

Question 99 of 129

x^2+y^2-2x+6y+6=0

(b) Evaluate:

\displaystyle\lim_{x\to 0} \dfrac{|x-1|+|x-2|-3}{2|x-1|-|x-2|}

Tamil Nadu DgeTamil Nadu HSC First Year (DGE) Board 2018Subjective· 5mImportance★★★★★

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Find each circle's centre and radius, compute the distance between the centres, and show it equals $|r_1-r_2|$ , which means the circles touch internally.

For a circle $x^2+y^2+2gx+2fy+c=0$ , the centre is $(-g,-f)$ and the radius is $\sqrt{g^2+f^2-c}$ .

Circle 1: $x^2+y^2-2x+6y+6=0 \Rightarrow 2g=-2,\ 2f=6,\ c=6 \Rightarrow g=-1,\ f=3$ .

Centre $C_1=(1,-3)$ , radius $r_1=\sqrt{(-1)^2+3^2-6}=\sqrt{1+9-6}=\sqrt{4}=2$ .

Circle 2: $x^2+y^2-5x+6y+15=0 \Rightarrow 2g=-5,\ 2f=6,\ c=15 \Rightarrow g=-2.5,\ f=3$ .

Centre $C_2=(2.5,-3)$ , radius $r_2=\sqrt{(-2.5)^2+3^2-15}=\sqrt{6.25+9-15}=\sqrt{0.25}=0.5$ .

Distance between centres:

$C_1C_2 = \sqrt{(2.5-1)^2+(-3-(-3))^2} = \sqrt{(1.5)^2+0} = 1.5$

Compare with the radii:

…

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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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Q.(a) Show that the circles $x^2+y^2-2x+6y+6=0$ and $x^2+y^2-5x+6y+15=0$ touch each other. OR

Concept understanding — General Equation of a Circle

Q.(a) Show that the circles x2+y2−2x+6y+6=0x^2+y^2-2x+6y+6=0x2+y2−2x+6y+6=0 and x2+y2−5x+6y+15=0x^2+y^2-5x+6y+15=0x2+y2−5x+6y+15=0 touch each other. OR

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Q.(a) Show that the circles $x^2+y^2-2x+6y+6=0$ and $x^2+y^2-5x+6y+15=0$ touch each other. OR

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