Test Index

Circle

Section: Mathematics

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Question : 2 of 89

Marks: +1, -0

If the equation of the circle passing through the point

(8,8)

and having the lines

x+2y-2=0

2x+3y-1=0

as its diameters is

x^2+y^2+px+qy+r=0

p^2+q^2+r=

[AP EAPCET 24 May 2025 S1]

$244$
$100$
$-44$
$44$

Solution:

Given, lines are

x+2y-2=0

and

2x+3y-1=0

Solving these two lines, we get

x=-4\;\text{and}\;y=3

Thus, the center of circle is

(-4,3)

.

Since, the general eq. of circle is

(x-h)^2+(y-k)^2=r^2

, where

(h,k)

is the center and

r

is the radius.

So,

(x-(-4))^2+(y-3)^2=r^2

\Rightarrow\;\;(x+4)^2+(y-3)^2=r^2\;\;\;\cdot\cdot\cdot\cdot\cdot\cdot\cdot(i)

But, the circle passes through the point

(8,8)

,

So,

(8+4)^2+(8-3)^2=r^2

12^2+5^2=r^2

169=r^2

\Rightarrow\;\;r=13

Now, from Eq. (i), we get

(x+4)^2+(y-3)^2=r^2\;=169

\Rightarrow\;\;x^2+8x+16+y^2-6y+9\;=169

\Rightarrow\;\;x^2+y^2+8x-6y-144\;=0

Comparing with the given equation

x^2+y^2+px+qx+r=0

We get

p=8,\;q=-6,\;r=-144

\therefore\;\;p^2+q^2+r\;=8^2+(-6)^2+(-144)

\;=64+36-144

\;=-44

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