Question

The number of common tangents to circle
x² + y² - 4x - 6y - 12 = 0 and
x² + y² + 6x + 18y + 26 = 0 is :

a) 1

b) 2

c) 3

d) 4

$\bf Note:$ Try To Use LaTex

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Answer 1

Answer:- Option (C) => 3

Explanation:-

(c) Central Idea Number of common tangents depends on the position of the circle with respects to each other.

(i) If circles touch externally.

${\sf\bold{{C_{1} }}}{\sf\bold{{C_{2} }}} = {\sf\bold{{r_{1} }}} + {\sf\bold{{r_{2} }}},3 \: {\sf\bold{{common \: tangents }}}</p><p></p><p>$

(ii) If circles touches internally

${\sf\bold{{C_{1} }}}{\sf\bold{{C_{2} }}} = {\sf\bold{{r_{2} }}} - {\sf\bold{{r_{1} }}},{\sf\bold{{1 \: common \: tangents }}}$

(iii) If circles do not touches each other,

4 common tangents.

Given Equations of circles are

x² + y² - 4x - 6y - 12 = 0 -------> (i)

x² + y² + 6x + 18y + 26 = 0 ---------> (ii)

$center \: of \: circles \: (i) \: is \:{\sf{{C_{1} }}} (2,3) \: and \: \\ radius.$

$= \sqrt{4 + 9 + 12} = 5({\sf\bold{{r_{1} }}} ) \: \: \: \: \: \: (say)$

$centre \: of \: circle \: (ii) \: is \: {\sf\bold{{C_{2} }}} ( - 3, - 9) \: and \: \\ radius.$

$= \sqrt{9 + 81 - 26} = 8({\sf\bold{{r_{2}) }}} \: \: \: \: \: \: (say)$

$Now, {\sf\bold{{C_{1} }}}{\sf\bold{{C_{2} }}} = \sqrt{(2 + 3) {}^{2} + (3 + 9) {}^{2} }$

${\sf\bold{{C_{1} }}}{\sf\bold{{C_{2} }}} = \sqrt{5 {}^{2} + 12 {}^{2} } = \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \sqrt{25 + 144 = 13}$

$Also, {\sf\bold{{r_{1} }}} + {\sf\bold{{r_{2} }}} = 5 + 8 = 13$

$\therefore {\sf\bold{{C_{1} }}} {\sf\bold{{C_{2} }}} = {\sf\bold{{r_{1} }}} + {\sf\bold{{r_{2} }}}$

Thus, both circles touch each other externally.

Hence, there are three common tangents.

Answer 2

Number of common tangents depend on the position of the circle with respect to the each other. (i) If circles touch externally C₁C₂ = ₁+ 2,3

common tangents

(ii) If circles touch internally → C₁ C₂ = 2-1, 1

common tangents (iii) If circles do not touch each other, 4 common tangents

Given equations of circles are

X₂ +y²-4x-6y-12 = 0 .. (1)

x²+y²+6x+18y+26 =0 ... (ii)

Centre of circle (i) is C1 (2,3) and

radius -√4+9+ 12 = 5 (r₁)

Centre of circle (ii) is C2(-3,-9) and

radius √9+81-26= 8 (r₂)

Now, C₁ C₂ = √(2+3)² + (3 + 9)²

➪ C₁ C₂ = √5²+12²

C₁ C₂ = √25+ 144 = 13

₁+r₂ = 5+ 8 = 13

Also, C₁ C₂ = ₁ +r₂

Thus, both circles touch each other externally. Hence, there are three common tangents.