The centre of a circle is $(2 a - 1, a) .$ Find the value of $a,$ if the circle passes through the point $(10, - 2)$ and has diameter $10 \sqrt{2}$ units.

Open in App

Solution

Here diameter
$\begin{matrix} 2 r = 10 \sqrt{2} r = 5 \sqrt{2} O A = r = 5 \sqrt{2} = \sqrt{(2 a - 1 - 10) + {(a + 2)}^{2}} = 5 \sqrt{2} {(2 a - 11)}^{2} + {(a + 2)}^{2} = 50 4 a^{2} - 44 a + 121 + a^{2} + 4 a + 4 = 50 5 a^{2} - 40 a + 75 = 0 a^{2} - 8 a + 15 = 0 a - 5 a - 3 a + 15 = 0 (a - 5 a) (a - 3) = 0 a = 5 o r 3 \end{matrix}$

Suggest Corrections

Similar questions

Q. Question 14
If the centre of a circle is (2a, a-7), then find the values of a , if the circle passes through the point (11,-9) and has diameter

10 \sqrt{2}

units.

Q. The centre of a circle is

(2 a + 3, 2 a - 1)

. Find the value of

a

if the circle passes through the point

(11, 9)

and has diameter of length

20

units.

Q. The centre of a circle is (

2 a, a - 7)

. Find the values of a if the circle passes through the point (11,

-

9) and has diameter

10 \sqrt{2}

units.

If the centre of a circle is (2a, a-7), then the values of 'a', if the circle passes through the point (11,-9) and has diameter $10 \sqrt{2}$ units will be:

Join BYJU'S Learning Program

The centre of a circle is (2a−1,a). Find the value of a, if the circle passes through the point (10,−2) and has diameter 10√2 units.

Here diameter2r=10√2r=5√2OA=r=5√2=√(2a−1−10)+(a+2)2=5√2(2a−11)2+(a+2)2=504a2−44a+121+a2+4a+4=505a2−40a+75=0a2−8a+15=0a−5a−3a+15=0(a−5a)(a−3)=0a=5or3

The centre of a circle is $(2 a - 1, a) .$ Find the value of $a,$ if the circle passes through the point $(10, - 2)$ and has diameter $10 \sqrt{2}$ units.