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Circle

Section: Mathematics

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Question : 10 of 89

Marks: +1, -0

If the point of contact of the circles

x^2+y^2-6x-4y+9=0

x^2+y^2+2x+2y-7=0

(\alpha, \beta)

7\beta=

[AP EAPCET 22 May 2025 S2]

$5\alpha$
$2\alpha$
$3\alpha$
$4\alpha$

Solution:

Circle I :

x^2+y^2-6x-4y+9=0

Circle II :

x^2+y^2+2x+2y-7=0

Point of contact :

(\alpha, \beta)

\Rightarrow (x-3)^2+(y-2)^2=4

Center

=(3,2)

(x+1)^2+(y+1)^2=9

centre

(-1,-1)

Slope of line

(m)=\frac{3}{4}

Equation of line

y-2=\frac{3}{4}(x-3)

\Rightarrow 3x-4y-1=0

The point of contact (

\alpha, \beta

) lies on the line

3x-4y-1=0

⇒ Radii are

r_1=2

and

r_2=3

Ratio

=2:3

Using section formula

\alpha=\frac{2(-1)+3(3)}{2+3}=\frac{7}{5}, \beta=\frac{2(-1)+3(2)}{2+3}

\Rightarrow \beta=\frac{4}{5} \Rightarrow 7\beta=7 \times \frac{4}{5}=4 \cdot \frac{7}{5}

\therefore 7\beta=4\alpha

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