You visited us 1 times! Enjoying our articles? Unlock Full Access!

Chord Properties of Circles

The equation ...

Question

The equation to the sides $A B, B C, C A$ of a $△$ are $x + y = 1; 4 x - y + 4 =$ and $2 x + 3 y = 6$ . Circle are drawn on $A B, B C, C A$ as diameter. The point of concurrence of the common chord is

centroid of the triangle

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

orthocenter

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

circumcenter

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

incenter

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

Open in App

Solution

The correct option is C orthocenter
Since $∠ A D B = ∠ A D C = 90^{0}$ , circle on $A B$ and $A C$ as diameter pass through $D$ and therefore the altitude $A D$ is the common chord.
Similarly, the other two common chords are the other two altitudes and hence they concur at the orthocenter.

Suggest Corrections

Similar questions

Q. The equation of sides

A B, B C, C A

of a triangle are

x + y = 1, 4 x - y + 4 = 0

and

2 x + 3 y = 6

then the equation of orthocentre through the vertex

A

Q. In

Δ A B C, A B = 3, B C = 4, C A = 5.

Semi circles are drawn outwardly on AB, BC as diameters. PQ is a common tangent to the semi circles. The length of PQ is

Three sides AB, BC and CA of a triangle ABC are $5 x - 3 y + 2 = 0, x - 3 y - 2 = 0$ and $x + y - 6 = 0$ respectively. Find the equation of the altitude through the vertex A.

Q. Three sides Ab, BC, and CA of a triangle ABC are

5 x - 3 y + 2 = 0, x - 3 y - 2, x + y - 6 = 0

respectively. Find the equations of the altitude through the vertex A.

Q. A second-degree curve inscribing the

Δ A B C,

where the sides

A B, B C

and

C A

have the equations

L_{1} L_{2} + λ L_{2} L_{3} + μ L_{3} L_{1} = 0,

where

λ

and

μ

are parameters. Let the equations of

A B, B C

and

C A

y = 2 x, y + 2 x = 0

x = 1

, respectively.

The circumcentre of the

Δ A B C

is:

Join BYJU'S Learning Program

The equation to the sides AB,BC,CA of a △ are x+y=1;4x−y+4= and 2x+3y=6. Circle are drawn on AB,BC,CA as diameter. The point of concurrence of the common chord is

The correct option is C orthocenterSince ∠ADB=∠ADC=900, circle on AB and AC as diameter pass through D and therefore the altitude AD is the common chord.Similarly, the other two common chords are the other two altitudes and hence they concur at the orthocenter.

The equation to the sides $A B, B C, C A$ of a $△$ are $x + y = 1; 4 x - y + 4 =$ and $2 x + 3 y = 6$ . Circle are drawn on $A B, B C, C A$ as diameter. The point of concurrence of the common chord is

The correct option is C orthocenter
Since $∠ A D B = ∠ A D C = 90^{0}$ , circle on $A B$ and $A C$ as diameter pass through $D$ and therefore the altitude $A D$ is the common chord.
Similarly, the other two common chords are the other two altitudes and hence they concur at the orthocenter.