In triangle $A B C$ , the equation of side $B C$ is $x - y = 0$ . Circumcentre and orthocentre of the triangle are $(2, 3)$ and $(5, 8)$ respectively. Equation of circumcircle of the triangle is

x^{2} + y^{2} - 4 x + 6 y - 27 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x^{2} + y^{2} - 4 x - 6 y - 27 = 0

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

x^{2} + y^{2} + 4 x + 6 y - 27 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x^{2} + y^{2} + 4 x - 6 y - 27 = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is D $x^{2} + y^{2} - 4 x - 6 y - 27 = 0$
Reflection of orthocenter $P (5, 8)$ in $B C (x - y = 0)$ will lie on circumcircle.
Clearly $P_{1} \equiv (8, 5)$ . Thus, the equation of circumcircle with center $O (2, 3)$ and passing through $P (8, 5)$ is
${(x - 2)}^{2} + {(y - 3)}^{2} = {(8 - 2)}^{2} + {(5 - 3)}^{2}$
i.e, $x^{2} + y^{2} - 4 x - 6 y - 27 = 0$

Suggest Corrections

Similar questions

Q. The equation of the image of the circle

x^{2} + y^{2} - 6 x - 4 y = 0

in the bisector of

2^{nd}

and

4^{th}

quadrant is

The diameters of a circle are along $2 x + y - 7 = 0$ and $x + 3 y - 11 = 0$ Then the equation of this circle, which also passes through (5,7) is

Tangents are drawn from the point P(1, 8) to the circle $x^{2} + y^{2} - 6 x - 4 y - 11 = 0$ touch the circle at the point A and B, then equation of the circumcircle of the triangle PAB is

Q. Tangents drawn from the point P(1, 8) to the circle

x^{2} + y^{2} - 6 x - 4 y - 11 = 0

touch the circle at points A and B. The equation of the circumcircle of triangle PAB is

Q. The equation of the circle concentric with the circle

x^{2} + y^{2} + 8 x + 10 y - 7 = 0

and passing through the centre of the circle

x^{2} + y^{2} - 4 x - 6 y = 0

Join BYJU'S Learning Program

In triangle ABC, the equation of side BC is x−y=0. Circumcentre and orthocentre of the triangle are (2,3) and (5,8) respectively. Equation of circumcircle of the triangle is

The correct option is D x2+y2−4x−6y−27=0Reflection of orthocenter P(5,8) in BC(x−y=0) will lie on circumcircle.Clearly P1≡(8,5). Thus, the equation of circumcircle with center O(2,3) and passing through P(8,5) is (x−2)2+(y−3)2=(8−2)2+(5−3)2i.e, x2+y2−4x−6y−27=0

In triangle $A B C$ , the equation of side $B C$ is $x - y = 0$ . Circumcentre and orthocentre of the triangle are $(2, 3)$ and $(5, 8)$ respectively. Equation of circumcircle of the triangle is