The two diameters of a circle are segments of the straight lines x-y=5 and 2 x+y=4. If the radius of the circle is 5 , then the equation of the circle is

Question

The two diameters of a circle are segments of the straight lines x-y=5 and 2 x+y=4. If the radius of the circle is 5 , then the equation of the circle is (A) x2+y2-6 x+4 y=12 (B) x2+y2-3 x+2 y=12 (C) x2+y2-6 x+2 y=12 (D) x2+y2-8 x+6 y-18=0

Tardigrade Admin · Accepted Answer

The intersecting point of lines x-y=5 and 2 x+y=4 is (3,-2) therefore Centre of circle is (3,-2). Required equation of circle is (x-3)2+(y+2)2=(5)2 ⇒ x2+9-6 x+y2+4+4 y=25 ⇒ x2+y2-6 x+4 y=12

Q. The two diameters of a circle are segments of the straight lines $x - y = 5$ and $2 x + y = 4$ . If the radius of the circle is 5 , then the equation of the circle is

$x^{2} + y^{2} - 6 x + 4 y = 12$

$x^{2} + y^{2} - 3 x + 2 y = 12$

$x^{2} + y^{2} - 6 x + 2 y = 12$

$x^{2} + y^{2} - 8 x + 6 y - 18 = 0$

$x^{2} + y^{2} - 8 x + 6 y - 7 = 0$

The intersecting point of lines $x - y = 5$
and $2 x + y = 4$ is $(3, - 2)$
therefore Centre of circle is $(3, - 2)$ .
Required equation of circle is
$(x - 3)^{2} + (y + 2)^{2} = (5)^{2}$
$\Rightarrow x^{2} + 9 - 6 x + y^{2} + 4 + 4 y = 25$
$\Rightarrow x^{2} + y^{2} - 6 x + 4 y = 12$

Q. The two diameters of a circle are segments of the straight lines x−y=5 and 2x+y=4. If the radius of the circle is 5 , then the equation of the circle is

x2+y2−6x+4y=12

x2+y2−3x+2y=12

x2+y2−6x+2y=12

x2+y2−8x+6y−18=0

x2+y2−8x+6y−7=0

The intersecting point of lines x−y=5 and 2x+y=4 is (3,−2) therefore Centre of circle is (3,−2). Required equation of circle is (x−3)2+(y+2)2=(5)2 ⇒x2+9−6x+y2+4+4y=25 ⇒x2+y2−6x+4y=12