A square is inscribed in a circle `x^2+y^2-6x+8y-103=0` such that its sides are parallel to co-ordinate axis then the distance of the nearest vertex to origin, is equal to

Options

A. 6
B. 13
C. `sqrt41`
D. `sqrt137`

Answer

C

Explanation

Given equation of circle is `x^(2)+y^(2)-6x + 8y - 103 = 0` which can be written as `(x-3)^(2)+(y+4)^(2) = 128 = (8sqrt2)^(2)`
`therefore` Centre = (3, -4) and radius = `8sqrt2`
Now, according to given information, we have the following figure.

for the coordinates of A and C.
Consider, `(x-3)/((1)/(sqrt2))=(y+4)/((1)/(sqrt2))=pm8sqrt2`
[using distance (parametric) from of line,
`(x-x_(1))/(costheta)=(y-y_(1))/(sintheta)=r`]
`rArr x=3pm8, y=-4pm8`
Similarly, for the coordinates of B and D, consider
`(x-3)/(-(1)/(sqrt2))=(y+4)/((1)/(sqrt2))=pm8sqrt2` [in this case, `theta= 135^(@)`]
`rArr c=3 pm8, y = -4 pm8`
`therefore B(11,-12)and D(-5, 4)`
`Now, OA=sqrt(25+144)=sqrt169=13`,
`OB=sqrt(121+144)=sqrt265`
`OC = sqrt(121+16)=sqrt41`
` and OD = sqrt(25+16)=sqrt41`