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Circle

Section: Mathematics

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Question : 3 of 89

Marks: +1, -0

The slope of one of the direct common tangents drawn to the circles

x^2+y^2-2x+4y+1=0

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

[AP EAPCET 23 May 2025 S2]

$0$
$\;\frac{4}{3}$
$\;\frac{3}{4}$
$1$

Solution:

\;c_1: x^2+y^2-2x+4y+1=0

\;\Rightarrow (x-1)^2+(y+2)^2=4

Centre

(c_1) =(1,-2), r_1=2

and

c_2: x^2+y^2-4x-2y+4=0

\Rightarrow (x-2)^2+(y-1)^2=1

Centre

(c_2) =(2,1), r_2=1

c_1 c_2=d={\sqrt{(2-1)^2+ (1+2^2.}}={\sqrt{10}}

Let the slope of direct common tangent be

m

the general line is

y=mx+c

this line has perpendicular distance

r_1

from

c_1

and

r_2

from

c_2

\; \therefore \left| \frac{m(1)-(-2)+c}{\sqrt{1+m^2}} \right|=2

\;\Rightarrow |m+2+c|=2 \sqrt{1+m^2}

\;\; \text{ and } \left| \frac{m(2)-1+c}{\sqrt{1+m^2}} \right|=1

\;\Rightarrow |2m+c-1|=\sqrt{1+m^2}

Put the value of '

c

' from Eq. (i) to Eq. (ii)

2m-1+(-m-2+2\sqrt{m^2+1}) = \sqrt{m^2+1}

\Rightarrow m=3-\sqrt{m^2+1}

Squaring both sides, we get

\;m^2=\;m^2+10

\Rightarrow \sqrt{m^2+1} -6\sqrt{m^2+1}

\Rightarrow\; m^2+1

\Rightarrow\; =\; \frac{25}{9}

\Rightarrow\; m^2=\frac{16}{9}

\Rightarrow\; m=\frac{4}{3}

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